获得 mathematica 用另一个方程来简化表达式 [英] Get mathematica to simplify expression with another equation

问题描述

我有一个非常复杂的 mathematica 表达式，我想通过使用一个新的、可能是无量纲的参数来简化它.

I have a very complicated mathematica expression that I'd like to simplify by using a new, possibly dimensionless parameter.

我表达的一个例子是:

K=a*b*t/((t+f)c*d);

(实际表达式非常大，有数千个字符).我想用 p

(the actual expression is monstrously large, thousands of characters). I'd like to replace all occurrences of the expression t/(t+f) with p

p=t/(t+f);

这里的目标是找到一个替代品，以便所有的 t 和 f 都被 p 替代.在这种情况下，替换 p 是一个无量纲化的参数，所以它看起来是一个很好的候选替换.

The goal here is to find a replacement so that all t's and f's are replaced by p. In this case, the replacement p is a nondimensionalized parameter, so it seems like a good candidate replacement.

我一直无法弄清楚如何在 mathematica 中做到这一点(或者如果可能的话).我试过了:

I've not been able to figure out how to do this in mathematica (or if its possible). I tried:

eq1= K==a*b*t/((t+f)c*d);
eq2= p==t/(t+f);
Solve[{eq1,eq2},K]

毫不奇怪，这行不通.如果有一种方法可以强制它以 p,a,b,c,d 的形式解决 K ，这可能会奏效，但我也不知道该怎么做.想法?

Not surprisingly, this doesn't work. If there were a way to force it to solve for K in terms of p,a,b,c,d, this might work, but I can't figure out how to do that either. Thoughts?

编辑 #1 (11/10/11 - 1:30)[删除以简化]

Edit #1 (11/10/11 - 1:30) [deleted to simplify]

好的，新机智.我采用 p=ton/(ton+toff) 并将 p 乘以几个表达式.我知道 p 可以完全消除.新的表达式(就 p 而言)是

OK, new tact. I've taken p=ton/(ton+toff) and multiplied p by several expressions. I know that p can be completely eliminated. The new expression (in terms of p) is

testEQ = A B p + A^2 B p^2 + (A+B)p^3;

然后我替换了 p，并调用了(正常的)FullSimplify，给出了这个表达式.

Then I made the substitution for p, and called (normal) FullSimplify, giving me this expression.

testEQ2= (ton (B ton^2 + A^2 B ton (toff + ton) + 
   A (ton^2 + B (toff + ton)^2)))/(toff + ton)^3;

最后，我尝试了下面的所有建议，除了最后一个(不确定它是如何工作的！)

Finally, I tried all of the suggestions below, except the last (not sure how it works yet!)

只有消除选项有效.所以我想我会从现在开始尝试这种方法.谢谢.

Only the eliminate option worked. So I guess I'll try this method from now on. Thank you.

EQ1 = a1 == (ton (B ton^2 + A^2 B ton (toff + ton) + 
        A (ton^2 + B (toff + ton)^2)))/(toff + ton)^3;
EQ2 = P1 == ton/(ton + toff);
Eliminate[{EQ1, EQ2}, {ton, toff}]

A B P1 + A^2 B P1^2 + (A + B) P1^3 == a1

我应该补充一点，如果目标是进行所有可能的替换，而剩下的，我仍然不知道该怎么做.但看起来如果替换可以完全消除一些变量，Eliminate[]效果最好.

I should add, if the goal is to make all substitutions that are possible, leaving the rest, I still don't know how to do that. But it appears that if a substitution can completely eliminate a few variables, Eliminate[] works best.

推荐答案

你试过了吗?

K = a*b*t/((t + f) c*d);
Solve[p == t/(t + f), t]
 -> {{t -> -((f p)/(-1 + p))}}

Simplify[K /. %[[1]] ]
 -> (a b p)/(c d)

哦，你知道消除?

Eliminate[{eq1, eq2}, {t,f}]
 -> a b p == c d K && c != 0 && d != 0

Solve[%, K]
 -> {{K -> (a b p)/(c d)}}

EDIT 2:此外，在这个简单的例子中，同时求解 K 和 t 似乎也能解决问题:

EDIT 2: Also, in this simple case, solving for K and t simultaneously seems to do the trick, too:

Solve[{eq1, eq2}, {K, t}]
 -> {{K -> (a b p)/(c d), t -> -((f p)/(-1 + p))}}

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