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Simplification of trigonometric expressions

You did not say what you expect the simplification to be. If you mean you wan to convert the input to 2 tans, you can make a rule ...
Nasser's user avatar
  • 142k
2 votes

Simplifying a logical expression with multivariate polynomials

The following is a very manual approach, but might be informative expr = a (a b + c d) != 0 || b (a b + c d) != 0 || c (a b + c d) != 0 || d (a b + c d) != 0; ...
mikado's user avatar
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2 votes

Simplifying a logical expression with multivariate polynomials

...
Bob Hanlon's user avatar
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4 votes

How can I get stronger simplification than FullSimplify?

1. First, let me note that one mainly applies the function FullSimplify if one deals with special functions. In application to algebraic expressions containing no ...
Alexei Boulbitch's user avatar
7 votes
Accepted

How to "factor out" units?

Mathematica can't cancel the ohms units because the unit dimensions of $b$ is an unknown quantity. However, if $b$ is defined as a "DimensionlessUnit" quantity, the ohms units will cancel. <...
creidhne's user avatar
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3 votes

How to "factor out" units?

This isn't exactly an answer to your question, but it may be a clue (I suspect that Mathematica won't simplify the expression because it doesn't know if a, b, c, and d are dimensionless quantities): <...
Cassini's user avatar
  • 5,556
4 votes

How to "factor out" units?

expr = Quantity[a, "Ohms"]/(b Quantity[1. c, "Ohms"] + Quantity[d, "Ohms"]); Using ReplaceAll <...
eldo's user avatar
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4 votes

Putting a term under the square root

You can add a transformation on top of the defaults that forces it: ...
Coolwater's user avatar
  • 20.1k
4 votes

Putting a term under the square root

A simple rule does it e1 = (1 + t)^2 Sqrt[1 + t^3/(1 + t)^3] e2 = e1 /. a_.*Sqrt[b_ + c_] :> Sqrt[ a^2*b + a^2*c] To see conditions when the above is valid, ...
Nasser's user avatar
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1 vote

Manipulating an expression into the form I want

Sometimes a specific rule is all what is needed to force what you want. ofcourse, this assumes the math is correct but since you said they are all positive, then this is OK. ...
Nasser's user avatar
  • 142k
2 votes

How to group terms of a Fourier series properly?

Try ...
Ulrich Neumann's user avatar
1 vote

Simplify recurrence relations with known solution form

Regarding the integral equation $$ Q_{n+1}(X) = \int_0^X Q_n (x) e^{k (X-x)} \ \mathrm{d} x, $$ it can be written as $$ Q_{n+1}(x) = Q_n(x)\circledast e^{kdx} $$ and taking the Laplace transform we ...
Cesareo's user avatar
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1 vote

Simplify recurrence relations with known solution form

Try NestList ...
Ulrich Neumann's user avatar
0 votes

How to group terms with the same exponential factor together after Expand?

Turn the comment to an answer: Collect[THGExp[t], Exp[_], Simplify]
3 votes

How do I replace certain parts of an expression

expr = 2 w E^(I w a/(v + c)) + 4 w^2 E^(I w a/(v - c)); p = Position[expr, E^x_] {{1, 2}, {2, 2}} Using ReplaceAt (new in 13.1)...
eldo's user avatar
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8 votes
Accepted

Why is Assuming[...] ignoring the assumption?

Assuming works by adding conditions to $Assumptions. Some functions make use of $Assumptions ...
Goofy's user avatar
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5 votes

Why is Assuming[...] ignoring the assumption?

...
azerbajdzan's user avatar
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13 votes
Accepted

Can Mathematica perform logical reasoning?

Just a slight variation on the excellent answer given by Bill, which is too long for a comment. Boolean Variables The truth values of gr, ...
gwr's user avatar
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16 votes

Can Mathematica perform logical reasoning?

This may be close. ...
Bill Watts's user avatar
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4 votes

Why doesn't Mathematica simplify $ie^{-\frac{i\pi}{2}(1+a+b)}$?

Instead of just posting workarounds, I will explain why they work. Simplify is a discrete, global minimization solver that minimizes a ...
Goofy's user avatar
  • 2,572
1 vote

Why doesn't Mathematica simplify $ie^{-\frac{i\pi}{2}(1+a+b)}$?

Another method with using ExpandAll prior to Simplify. ...
Domen's user avatar
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1 vote

Why doesn't Mathematica simplify $ie^{-\frac{i\pi}{2}(1+a+b)}$?

I guess there is a limit to what Simplify can do. Here is another alternative expr = I*Exp[-I Pi/2*(1 + a + b)]; Simplify@TrigToExp@Simplify@ExpToTrig@expr
Nasser's user avatar
  • 142k
4 votes

Why doesn't Mathematica simplify $ie^{-\frac{i\pi}{2}(1+a+b)}$?

I * Exp[-I Pi / 2 * (1 + a + b)] // ComplexExpand // FullSimplify E^(-(1/2) I (a+b) [Pi])
AsukaMinato's user avatar
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5 votes
Accepted

Making Mathematica return the full simplified expression by default

How about taking the limit instead: ...
Nasser's user avatar
  • 142k
1 vote

FullSimplify exponential gives zero

It is a numerical underflow, but unfortunately without a concomitant message. If you use software reals instead of machine reals, you get the right answer: ...
Glenn Welch's user avatar
2 votes

FullSimplify exponential gives zero

I am on the side of a bug, although as @Daniel Lichtblau points out in the comment: In presence of machine numbers and operations such as exponentiation one can get an underflow to zero. but no ...
azerbajdzan's user avatar
  • 13.4k
2 votes

How can I expand the inner parentheses of trigonometric functions?

To keep Cos instead of turning into Sin: ...
AccidentalFourierTransform's user avatar
4 votes
Accepted

How can I expand the inner parentheses of trigonometric functions?

Use ExpandAll. However, it will also turn Cos into Sin. ...
Domen's user avatar
  • 21.9k
1 vote
Accepted

Transforming expression containing powers of parameter

You've not given us a lot of test cases, so here is an implementation with FactorList that works for your example. ...
Domen's user avatar
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