Re: Simplification de formule/ Simplification of formula
- To: mathgroup at smc.vnet.net
- Subject: [mg124477] Re: Simplification de formule/ Simplification of formula
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Sat, 21 Jan 2012 05:14:11 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jf696v$gdp$1@smc.vnet.net>
Reverse is not limited to reversing lists:
r = {a -> b}; s = {a -> b, c -> d};
Reverse /@ r
{b -> a}
Reverse /@ s
{b -> a, d -> c}
Bob Hanlon
On Fri, Jan 20, 2012 at 1:52 AM, Dr. Wolfgang Hintze <weh at snafu.de> wrote:
> Here's a possible solution to the problem of inverting replacement
> rules.
>
>
> In[42]:=
> r = {a -> b}; s = {a -> b, c -> d};
> In[40]:=
> (Rule @@ Sequence[Reverse[List @@ #1]] & ) /@ r
> Out[40]=
> {b -> a}
> In[41]:=
> (Rule @@ Sequence[Reverse[List @@ #1]] & ) /@ s
> Out[41]=
> {b -> a, d -> c}
>
> For a "naked" rule (not in a list) you should use
>
> In[43]:=
> t = u -> v;
> (Rule @@ Sequence[Reverse[List @@ #1]] & )[t]
> Out[44]=
> v -> u
>
> Regards,
> Wolfgang
>
> "Dr. Wolfgang Hintze" <weh at snafu.de> schrieb im Newsbeitrag
> news:jf8qd2$3ld$1 at smc.vnet.net...
>>> I want to simplify this formula :
>>
>> This is obviously a good idea and a good excercise if it comes to
>> obtaining an overview over a confusing looking expression.
>>
>> You cannot expect fully automatic simplification, since it was not
>> clearly stated which should be the desired form.
>> So it will be a dialog between you and Mathematica.
>>
>> I suggest to first to make some replacements to get better
>> readability
>> (you can undo them in the end).
>> Then recognising the structure by eye it will almost become obvious
>> which commands to choose.
>>
>> Ok, let's do it!
>>
>> 1) Let your expression be
>>
>> x = -((DK*Pv2*Cos[a2 - alpha3]*Cos[aV2]*Sin[a2 - aV2])/(DK*Cos[a2 -
>> alpha3] - KF*Sin[a2 - alpha3])) +
>> (KF*Pv2*Cos[aV2]*Sin[a2 - alpha3]*Sin[a2 - aV2])/(DK*Cos[a2 -
>> alpha3] -
>> KF*Sin[a2 - alpha3]) -
>> (DG3*P3*Sin[alpha3]*Sin[a2 - aV2])/(DK*Cos[a2 - alpha3] - KF*Sin[a2 -
>> alpha3]) - (DK*FV2*Pv2*Cos[a2 - alpha3]*Cos[a2 - aV2]*Sin[aV2])/
>> (L2*(DK*Cos[a2 - alpha3] - KF*Sin[a2 - alpha3])) +
>> (FV2*KF*Pv2*Cos[a2 -
>> aV2]*Sin[a2 - alpha3]*Sin[aV2])/
>> (L2*(DK*Cos[a2 - alpha3] - KF*Sin[a2 - alpha3])) - (KF*Pv2*Cos[a2 -
>> alpha3]*Sin[a2 - aV2]*Sin[aV2])/(DK*Cos[a2 - alpha3] - KF*Sin[a2 -
>> alpha3]) +
>> (FV2*KF*Pv2*Cos[a2 - alpha3]*Sin[a2 - aV2]*Sin[aV2])/(L2*(DK*Cos[a2 -
>> alpha3] - KF*Sin[a2 - alpha3])) -
>> (DK*Pv2*Sin[a2 - alpha3]*Sin[a2 - aV2]*Sin[aV2])/(DK*Cos[a2 -
>> alpha3] -
>> KF*Sin[a2 - alpha3]) + (DK*FV2*Pv2*Sin[a2 - alpha3]*Sin[a2 -
>> aV2]*Sin[aV2])/
>> (L2*(DK*Cos[a2 - alpha3] - KF*Sin[a2 - alpha3]))
>>
>> 2) Replacements
>>
>> x1 = x /. {a2 -> a, alpha3 -> b, aV2 -> c, DK -> A, Pv2 -> B, KF ->
>> U,
>> DG3 -> V, P3 -> W, FV2 -> X, L2 -> Y}
>>
>> -((A*B*Cos[a - b]*Cos[c]*Sin[a - c])/(A*Cos[a - b] - U*Sin[a - b])) +
>> (B*U*Cos[c]*Sin[a - b]*Sin[a - c])/(A*Cos[a - b] - U*Sin[a - b]) -
>> (V*W*Sin[b]*Sin[a - c])/(A*Cos[a - b] - U*Sin[a - b]) -
>> (A*B*X*Cos[a -
>> b]*Cos[a - c]*Sin[c])/(Y*(A*Cos[a - b] - U*Sin[a - b])) +
>> (B*U*X*Cos[a - c]*Sin[a - b]*Sin[c])/(Y*(A*Cos[a - b] - U*Sin[a -
>> b])) - (B*U*Cos[a - b]*Sin[a - c]*Sin[c])/(A*Cos[a - b] - U*Sin[a -
>> b])
>> +
>> (B*U*X*Cos[a - b]*Sin[a - c]*Sin[c])/(Y*(A*Cos[a - b] - U*Sin[a -
>> b])) - (A*B*Sin[a - b]*Sin[a - c]*Sin[c])/(A*Cos[a - b] - U*Sin[a -
>> b])
>> +
>> (A*B*X*Sin[a - b]*Sin[a - c]*Sin[c])/(Y*(A*Cos[a - b] - U*Sin[a -
>> b]))
>>
>> By the way, this expression looks much nicer in Mathematica than in
>> this mail here.
>>
>> 3) Seeing the common denominator we tell Mathematica to extract it
>> x2 = Collect[x1, A*Cos[a - b] - U*Sin[a - b]]
>>
>> (1/(A*Cos[a - b] - U*Sin[a - b]))*((-A)*B*Cos[a - b]*Cos[c]*Sin[a -
>> c]
>> + B*U*Cos[c]*Sin[a - b]*Sin[a - c] - V*W*Sin[b]*Sin[a - c] -
>> (A*B*X*Cos[a - b]*Cos[a - c]*Sin[c])/Y + (B*U*X*Cos[a - c]*Sin[a -
>> b]*Sin[c])/Y - B*U*Cos[a - b]*Sin[a - c]*Sin[c] +
>> (B*U*X*Cos[a - b]*Sin[a - c]*Sin[c])/Y - A*B*Sin[a - b]*Sin[a -
>> c]*Sin[c] + (A*B*X*Sin[a - b]*Sin[a - c]*Sin[c])/Y)
>>
>> Again this expression looks much nicer in Mathematica than in this
>> mail
>> here.
>>
>> 4) Trying to extrax 1/Y using collect didn't improve things so we try
>> a
>> Simplify in the end
>>
>> x3 = Simplify[x2]
>>
>> -((1/(2*Y*(A*Cos[a - b] - U*Sin[a - b])))*(B*U*X*Cos[2*a - b] -
>> B*U*Y*Cos[b] - B*U*X*Cos[2*a - b - 2*c] + B*U*Y*Cos[2*a - b - 2*c] +
>> V*W*Y*Cos[a - b - c] -
>> V*W*Y*Cos[a + b - c] + A*B*X*Sin[2*a - b] + A*B*Y*Sin[b] -
>> A*B*X*Sin[2*a - b - 2*c] + A*B*Y*Sin[2*a - b - 2*c]))
>>
>> This seems to be a rather good result:
>>
>> - one common denominator extracted
>> - all terms in the big bracket have the same structure A*B*C
>> Trig[...]
>> - angles collected under the Trigfunctions
>>
>> PS: here is a little additional exercise: how would you invert the
>> replacements 2) efficiently?
>>
>> Hope this helps
>> Best regards,
>> Wolfgang
>>
>>
>> "Nicolas Simon" <nicodumonastal at gmail.com> schrieb im Newsbeitrag
>> news:jf696v$gdp$1 at smc.vnet.net...
>>>
>>> Hi !
>>>
>>> I want to simplify this formula :
>>>
>>> -((DK Pv2 Cos[a2 - alpha3] Cos[aV2] Sin[a2 - aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) + (
>>> KF Pv2 Cos[aV2] Sin[a2 - alpha3] Sin[a2 - aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) - (
>>> DG3 P3 Sin[alpha3] Sin[a2 - aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) - (
>>> DK FV2 Pv2 Cos[a2 - alpha3] Cos[a2 - aV2] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) + (
>>> FV2 KF Pv2 Cos[a2 - aV2] Sin[a2 - alpha3] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) - (
>>> KF Pv2 Cos[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) + (
>>> FV2 KF Pv2 Cos[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) - (
>>> DK Pv2 Sin[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) + (
>>> DK FV2 Pv2 Sin[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])).
>>>
>>> By manually regrouping the terms, the denominator disappears for a
>>> large number of fraction but I'm not able to reproduce this result
>>> with Simplify or others Mathematica tools to handle formula. Any
>>> suggestion of how I could do that with Mathematica ?
>>> Thank you for your help !
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>> Bonjour,
>>>
>>> Je souhaite simplifier cette formule :
>>>
>>> -((DK Pv2 Cos[a2 - alpha3] Cos[aV2] Sin[a2 - aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) + (
>>> KF Pv2 Cos[aV2] Sin[a2 - alpha3] Sin[a2 - aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) - (
>>> DG3 P3 Sin[alpha3] Sin[a2 - aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) - (
>>> DK FV2 Pv2 Cos[a2 - alpha3] Cos[a2 - aV2] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) + (
>>> FV2 KF Pv2 Cos[a2 - aV2] Sin[a2 - alpha3] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) - (
>>> KF Pv2 Cos[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) + (
>>> FV2 KF Pv2 Cos[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])) - (
>>> DK Pv2 Sin[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3]) + (
>>> DK FV2 Pv2 Sin[a2 - alpha3] Sin[a2 - aV2] Sin[aV2])/(
>>> L2 (DK Cos[a2 - alpha3] - KF Sin[a2 - alpha3])).
>>>
>>> En regroupant les bons termes le d=E9nominateur se simplifie pour un
>>> grand nombres de fractions mais je n'arrive pas a ce r=E9sultat avec
>>> les
>>> outils de simplification (FullSimplify ou Simplify) ou les outils de
>>> manipulations de formule. Est ce que vous auriez une id=E9e de
>>> comment
>>> faire comme pr=E9ciser certains "assumption" dans Simplify ?
>>> Merci de votre aide.
>>>