I'm using SBML which is an XML format for capturing numerical simulations. This uses MathML to capture equations that relate the amounts of the different things in the simulation. When I write the equations down by hand, I add some brackets that are not strictly needed, but which make the equations easier to read. For example, I may have an equation that by hand I'd write as:
z = (ax + by) + (cx + dy)
When this goes into MathML, these brackets are stripped out, and once I pretty-print the MathML, it comes out as:
z = ax + by + cx + dy
This makes me sad. Is there a way to capture these (redundant) brackets in the MathML expression?
Yes, you don't say if you are using presentation or content mathml. I would guess content as it would be rare for you to lose the brackets in presentation rendering.
Content MathML you could mark it up as
<apply><sum/>
<apply><csymbol>bracket</csymbol>
<apply><sum/>
<apply><product/><mi>a</mi><mi>x</mi></apply>
<apply><product/><mi>b</mi><mi>y</mi></apply>
</apply>
</apply>
<apply><csymbol>bracket</csymbol>
<apply><sum/>
<apply><product/><mi>c</mi><mi>x</mi></apply>
<apply><product/><mi>d</mi><mi>y</mi></apply>
</apply>
</apply>
</apply>
where you give the csymbol bracket a presentation rule of putting a bracket round its argument.
Related
I'd like something to convert pretty basic math expressions, having nested parentheses and fractions, to LaTeX notation. Like mathquill, but a function (or even the building blocks of one).
There seem to be some Lua and Haskell solutions in a pandoc/Rmarkdown context, but I can't use those, because (a) I'm scared of real languages, and (b) I'm generating PNGs (via webtex) to be featured in a flextable table, outside of a rendered document.
I'm inexperienced with regular expressions, so I don't know how to leverage something like this, but I'd appreciate any pointers if that seems like a productive path.
"Write a parser" is something best left to others, at least in my case!
Example expression below. Just a few levels of nesting, no big deal. I can bound it at, say, 5 levels, if that helps. And the input is parseable as an R expression—I can guarantee matched parentheses, for example.
(y0 - (y0/((1-y0)*exp(B*dx)+y0)))*Pop
Here's what I'd want in the case above. The \cdots are sugar; I can handle those.
\left(y0-\frac{y0}{\left(1-y0\right)\cdot\exp\left(B\cdot dx\right)+y0}\right)\cdot Pop
Visually:
I keep encountering an awkward problem when I type out expressions in LaTeX where the text line in the compiled document cuts an expression/equation/math object in half. Of course, I make use of $$ $$ tags to ensure this doesn't happen for particularly hefty expressions or computations, but this isn't appropriate for small expressions that appear in expository text.
Is there something I can do to make sure LaTeX never chops a mathmode expression in half in the compiled document?
you can use \mbox around your equation
\mbox{$E = mc^2$}
This will prevent a line break in the equation if it is located close to the max text width, but may lead to the equation reaching over the end of the text body.
Here is a simple solution just appending a tilde ~ to equations.
Example borrowed from here
Instead of
$x = \{x\,|\,x\in\mathbb{R_{\ge0}}\}$
you would write
$~{x = \{x\,|\,x\in\mathbb{R_{\ge0}}\}}$
I like solving my math problems(high school) using R as it is faster than writing on a piece of paper. One problem I'm having is that I have to keep writing the multiplication sign, example:
9x^2 + 24x + 16 yields = Error: unexpected symbol in "9x"
Is there any way in R to multiply 4x, without having to write 4*x but only 4x?
Would save me some time in having to write one extra character the whole time! Thanks
No. Having a number in front of a character without any space simply isn't valid syntax in R.
Take a step back and look at the syntax rules for, say, Excel, Matlab, Python, Mathematica. Every language has its rules, generally (:-) ) with good reason. For example, in R, the following are legal object names:
foo
foo.bar
foo1
foo39
But 39foo is not legal. So if you wanted any sequence [0-9][Letters] or the reverse to indicate multiplication, you'd have a conflict with naming rules.
how to write a formula like
v_r (t)=∑_(n=0)^(N-1)▒〖A_r (L_2-L_1 ) e^j(ω_c t-4π/λ (R+υt+L_(1+L_2 )/2 cos〖(θ)sin(ω_r t+2πn/N)))〗 ┤) sinc(4π/λ-L_(2-L_1 )/2 cos(θ) sin(ω_r t+2πn/N))〗
in c#?
You have to convert the formula to something the compiler recognizes.
To it's equivalent using the a combination of basic algebra and the Math class like so:
p = rho*R*T + (B_0*R*T-A_0-((C_0) / (T*T))+((E_0) / (Math.Pow(T, 4))))*rho*rho +
(b*R*T-a-((d) / (T)))*Math.Pow(rho, 3) +
alpha*(a+((d) / (t)))*Math.Pow(rho, 6) +
((c*Math.Pow(rho, 3)) / (T*T))*(1+gamma*rho*rho)*Math.Exp(-gamma*rho*rho);
Example taken from: Converting Math Equations in C#
Well, first you have to figure out what all those symbols mean. I see the sigma which usually indicates sum-of, with ∑_(n=0)^(N-1) probably translating to:
N-1
∑
n=0
This generally means the sum of the following expression where n varies from 0 to N-1. So I gather you'd need a loop there.
The expression to be calculated within that loop consists of a lof of trigonometric-type functions involving π, θ, sin and cos, and the little known sinc which I assume is a typo :-)
The bottom line is that you need to understand the current expression before you can think about converting it to another form like a C# program. Short of knowing where it came from, or a little bit of context, we probably can't help you that much though there's always a possibility that we have a savant/genius here that will recognise that formula off the top of their head.
I'm interested in building a derivative calculator. I've racked my brains over solving the problem, but I haven't found a right solution at all. May you have a hint how to start? Thanks
I'm sorry! I clearly want to make symbolic differentiation.
Let's say you have the function f(x) = x^3 + 2x^2 + x
I want to display the derivative, in this case f'(x) = 3x^2 + 4x + 1
I'd like to implement it in objective-c for the iPhone.
I assume that you're trying to find the exact derivative of a function. (Symbolic differentiation)
You need to parse the mathematical expression and store the individual operations in the function in a tree structure.
For example, x + sin²(x) would be stored as a + operation, applied to the expression x and a ^ (exponentiation) operation of sin(x) and 2.
You can then recursively differentiate the tree by applying the rules of differentiation to each node. For example, a + node would become the u' + v', and a * node would become uv' + vu'.
you need to remember your calculus. basically you need two things: table of derivatives of basic functions and rules of how to derivate compound expressions (like d(f + g)/dx = df/dx + dg/dx). Then take expressions parser and recursively go other the tree. (http://www.sosmath.com/tables/derivative/derivative.html)
Parse your string into an S-expression (even though this is usually taken in Lisp context, you can do an equivalent thing in pretty much any language), easiest with lex/yacc or equivalent, then write a recursive "derive" function. In OCaml-ish dialect, something like this:
let rec derive var = function
| Const(_) -> Const(0)
| Var(x) -> if x = var then Const(1) else Deriv(Var(x), Var(var))
| Add(x, y) -> Add(derive var x, derive var y)
| Mul(a, b) -> Add(Mul(a, derive var b), Mul(derive var a, b))
...
(If you don't know OCaml syntax - derive is two-parameter recursive function, with first parameter the variable name, and the second being mathched in successive lines; for example, if this parameter is a structure of form Add(x, y), return the structure Add built from two fields, with values of derived x and derived y; and similarly for other cases of what derive might receive as a parameter; _ in the first pattern means "match anything")
After this you might have some clean-up function to tidy up the resultant expression (reducing fractions etc.) but this gets complicated, and is not necessary for derivation itself (i.e. what you get without it is still a correct answer).
When your transformation of the s-exp is done, reconvert the resultant s-exp into string form, again with a recursive function
SLaks already described the procedure for symbolic differentiation. I'd just like to add a few things:
Symbolic math is mostly parsing and tree transformations. ANTLR is a great tool for both. I'd suggest starting with this great book Language implementation patterns
There are open-source programs that do what you want (e.g. Maxima). Dissecting such a program might be interesting, too (but it's probably easier to understand what's going on if you tried to write it yourself, first)
Probably, you also want some kind of simplification for the output. For example, just applying the basic derivative rules to the expression 2 * x would yield 2 + 0*x. This can also be done by tree processing (e.g. by transforming 0 * [...] to 0 and [...] + 0 to [...] and so on)
For what kinds of operations are you wanting to compute a derivative? If you allow trigonometric functions like sine, cosine and tangent, these are probably best stored in a table while others like polynomials may be much easier to do. Are you allowing for functions to have multiple inputs,e.g. f(x,y) rather than just f(x)?
Polynomials in a single variable would be my suggestion and then consider adding in trigonometric, logarithmic, exponential and other advanced functions to compute derivatives which may be harder to do.
Symbolic differentiation over common functions (+, -, *, /, ^, sin, cos, etc.) ignoring regions where the function or its derivative is undefined is easy. What's difficult, perhaps counterintuitively, is simplifying the result afterward.
To do the differentiation, store the operations in a tree (or even just in Polish notation) and make a table of the derivative of each of the elementary operations. Then repeatedly apply the chain rule and the elementary derivatives, together with setting the derivative of a constant to 0. This is fast and easy to implement.