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8 votes
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Express in forms of a

We can solve y and eliminate x. ...
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7 votes
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Automatic identification and substitution of fundamental mathematical constants

How can I computationally force or convert or express results to be expressed in mathematical constants, if possible? For the GoldenRatio, it shows up as $\phi$ only in TraditionalForm. So it is not ...
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1 vote

How can I reduce this complex exponent expression to real expression?

The last two terms can be simplified to real as follows FullSimplify[-(c - d I) E^((a + b I) t) - (c + d I) E^((a - b I) t)==-2 E^(a t) (c Cos[b t] + d Sin[b t])]/. ...
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  • 11
1 vote

How can I reduce this complex exponent expression to real expression?

This is a problem created by using machine numbers. These are seldom 100% accurate, they are most of the time an approximation. And if you have an expression like: x - Conjugate[x] it is possible that ...
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3 votes
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How can I reduce this complex exponent expression to real expression?

Rationalize[expr, 0] // Im // ComplexExpand 0. It means that expr is real. ...
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10 votes

Why Mathematica is treating the product of (specified) real variable as complex?

You are probably not using Mathematica correctly. First, you shouldn't put MatrixForm into SingularValueList. Second, the way ...
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9 votes
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How to write an easy Conjugate Function that just replaces i -> -i

Replace acts on the full form of an expression, not the displayed form. The full form of your expression: expr = I (a + 0.5123 I b - 1.2332 I c); expr // FullForm Therefore a complex number is ...
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2 votes
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How to make wolfram simplify sqrt[x^2] when x > 0?

...
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2 votes

Distribute conjugate and simplify to absolute values

You could define your own "fake" symbolic conjugate that has the properties you want and then use the real Conjugate only when you need it. ...
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2 votes
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Distribute conjugate and simplify to absolute values

Not to make things too complicated, I only use the first part of w for an example: w = 1/2 (x Conjugate[x] + y Conjugate[y] - a x Conjugate[-a x + a y]) Then ...
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2 votes

Factorize in a particular way

A quick way to do that is: Collect[TestPolynomial,{x1,x2,x3},Simplify] An inconvenience with the above code is that it also factorizes some of the variables : ...
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2 votes
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Mathematica won't validate Fresnel Integral identity

Here is OP's formula from his Out[40] stored in expression: ...
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7 votes

Mathematica won't validate Fresnel Integral identity

Works fine here once I correct the formula: FresnelC[z] + I FresnelS[z] == (1 + I)/2 Erf[(1 - I)/2 Sqrt[π] z] // FullSimplify (* True *) To get the original ...
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2 votes

Square root of complex exponenial does not simplify

which should simplify to simple (1+2Nph) One way to simplify this is to to simplify with side relation ...
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2 votes
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Reduce and FullSimplify don't reduce fully -- how to check if candidate solutions hold

Simplify[ x a^2 + y b^2 + (-x - y) a*b == 0, {a == 1/x, b == 1/y, y > x > 1}] True.
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3 votes

Reduce and FullSimplify don't reduce fully -- how to check if candidate solutions hold

Provide the conditions to FullSimplify too. ...
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  • 16.5k
1 vote

Simplifying DSolve output. Exponentials raised to constant

With the solutions, DSolve gives, you can access all real solutions for real x with real c1. Substituting E^c1 -> c2 and restricting c2 also to real values, you ...
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4 votes

Solve conditional term with mathematica

We consider that a[n]=x^n + y^n. To deduce the recurrence relation we compare with x^(n + 2) + y^(n + 2) and ...
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6 votes

Solve conditional term with mathematica

A simpler way is as follows. Simplify[(x^3 + y^3)/(x^2 + y^2), {x + y == 3, x y == -1}] 36/11
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8 votes

Solve conditional term with mathematica

The lazy way to do this is to evaluate Solve[{z == (x^3 + y^3)/(x^2 + y^2), x + y == 3, x y == -1}, z, {x, y}] {{z -> 36/11}} The clever way to do this is ...
2 votes

Solve conditional term with mathematica

One way could be to first solve for $x,y$ using Solve command, and then replace all the solutions into the expression. Something like ...
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  • 113k
3 votes
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Simple way to express disjunction likes x==a || x==b || x==c ||

Not being an expert in the statistical functions, I'm not sure about this, but would something like this work? ...
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