Group Abstract

Message Boards

WOLFRAM COMMUNITY

6.2K Views

7 Replies

2 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Mathematics Equation Solving Wolfram Language

Why does WL not know that Log[a]-Log[b] is Log[a/b]?

Werner Geiger

Posted 8 years ago

POSTED BY: Werner Geiger

7 Replies

Sort By:

Werner Geiger

Posted 8 years ago

BTW: To avoid the ConditionalExpression from Integrate, which is useless in my case, one can suppress them with option GenerateConditions. Instead auf writing: In[359]:= v[s_] := m s + c; t5 = Assuming[Reals, Integrate[1/v[x], {x, s0, s}]] Out[360]= ConditionalExpression[(Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m, s0 > 0 && Re[s] > s0 && s == Re[s]] ... one can write: In[361]:= v[s_] := m s + c; t6 = Assuming[Reals, Integrate[1/v[x], {x, s0, s}, GenerateConditions -> False]] Out[362]= (Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m

BTW: To avoid the ConditionalExpression from Integrate, which is useless in my case, one can suppress them with option GenerateConditions. Instead auf writing:

In[359]:= v[s_] := m s + c;
t5 = Assuming[Reals, Integrate[1/v[x], {x, s0, s}]]

Out[360]= ConditionalExpression[(Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m, 
s0 > 0 && Re[s] > s0 && s == Re[s]]

... one can write:

In[361]:= v[s_] := m s + c;
t6 = Assuming[Reals, 
  Integrate[1/v[x], {x, s0, s}, GenerateConditions -> False]]

Out[362]= (Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m

POSTED BY: Werner Geiger

Hans Milton

Posted 8 years ago

A kind of roundabout way: In[1]:= Log@Exp[Log@a - Log@b] Out[1]= Log[a/b] Applied to Werners example: In[2]:= t = (Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m Out[2]= (Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m In[3]:= t /. Log[a_] - Log[b_] -> Log@Exp[Log@a - Log@b] Out[3]= Log[(1 + (m s)/c)/(1 + (m s0)/c)]/m

A kind of roundabout way:

In[1]:= Log@Exp[Log@a - Log@b]

Out[1]= Log[a/b]

Applied to Werners example:

In[2]:= t = (Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m

Out[2]= (Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m

In[3]:= t /. Log[a_] - Log[b_] -> Log@Exp[Log@a - Log@b]

Out[3]= Log[(1 + (m s)/c)/(1 + (m s0)/c)]/m

POSTED BY: Hans Milton

Werner Geiger

Posted 8 years ago

Your Replace-idea will certainly work and does not disturb anything if my expression does not contain loag(a)-log(b). I think it can even be simplified: In[275]:= v[s_] := m s + c; t3 = Assuming[{Reals, v[x] > 0}, Integrate[1/v[x], {x, s0, s}]] Out[276]= ConditionalExpression[(Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m, s0 > 0 && Re[s] > s0 && s == Re[s]] In[277]:= t3 = Simplify[t3 /. Log[a_] - Log[b_] -> Log[a/b]] Out[277]= ConditionalExpression[Log[(c + m s)/(c + m s0)]/m, s0 > 0 && Re[s] > s0 && s == Re[s]]

Your Replace-idea will certainly work and does not disturb anything if my expression does not contain loag(a)-log(b). I think it can even be simplified:

In[275]:= v[s_] := m s + c;
t3 = Assuming[{Reals, v[x] > 0}, Integrate[1/v[x], {x, s0, s}]]

Out[276]= ConditionalExpression[(Log[1 + (m s)/c] - Log[1 + (m s0)/c])/m, 
 s0 > 0 && Re[s] > s0 && s == Re[s]]

In[277]:= t3 = Simplify[t3 /. Log[a_] - Log[b_] -> Log[a/b]]

Out[277]= ConditionalExpression[Log[(c + m s)/(c + m s0)]/m, 
 s0 > 0 && Re[s] > s0 && s == Re[s]]

POSTED BY: Werner Geiger

Frank Kampas

Frank Kampas, Physicist at Large Consulting

Posted 8 years ago

In[3]:= Reduce[Log[a] - Log[b] == Log[a/b], {a, b}, Reals] Out[3]= a > 0 && b > 0

POSTED BY: Frank Kampas

Werner Geiger

Posted 8 years ago

Ja, of course. That's known, but not my problem.

POSTED BY: Werner Geiger

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 8 years ago

(1) I am pretty sure `Assuming[Reals,...]` has no meaning. `Reals` can be used as a domain in solvers that accept such an argument. `Assumptions` by contrast expects explicit assumptions on stated parameters e.g. `c>0`. (2) That second `Simplify` looks fine, in that log a - log b is not equal to log a/b for arbitrary real (a,b). Log[a] - Log[b] - Log[a/b] /. {a -> 3, b -> -1} (* Out[1]= -2 I \[Pi] ) (3) For parametrized definite integrals, it generally helps to use as many assumptions as possible. Below is an example which could be modified for specific needs. v[s_] := m s + c; t1 = Integrate[1/v[x], {x, s0, s}, Assumptions -> {m > 0, c > 0, 0 < s < s0}] ( Out[416]= Log[(c + m s)/(c + m s0)]/m *)

POSTED BY: Daniel Lichtblau

Werner Geiger

Posted 8 years ago

Thanks, Daniel. Of course Log[a]-Log[b}==Log[a/b holds true only for a>0 && b>0. But this is not my problem. Your proposition (3) doesn't help since it makes assumptions about the structure of my expression. Actually I do not know anything about Integrate[1/v[x],{x,s0,s}] apart from v[x] being real and positive within the real integration interval [s0,s]. Not even s>=s0 needs to be. Hence I think I cannot write more than the t2 in my original post. Anyway I can live with log(a)-log(b) not being resolved to log(a/b).

POSTED BY: Werner Geiger

Reply to this discussion

Reply Preview

Attachments

Remove Add a file to this post

Follow this discussion

or Discard

Feedback