mathematical | Convert mathematical equations to SVGs , PNGs , or MathML | Math library
kandi X-RAY | mathematical Summary
kandi X-RAY | mathematical Summary
Quickly convert math equations into beautiful SVGs (or PNGs/MathML).
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of mathematical
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mathematical Examples and Code Snippets
function rs(){ql.startNonterminal("FunctionItemExpr",Ll);switch(Al){case 145:El(92);break;default:Cl=Al}switch(Cl){case 32:case 17553:us();break;default:ss()}ql.endNonterminal("FunctionItemExpr",Ll)}
Community Discussions
Trending Discussions on mathematical
QUESTION
I have a string formed by number and mathematical operator like "1 + 1 *1"
that is the text content of the number appendend on the screen div, I want to form an array of them and then divide it using mathematical operators such as + or - as a divisor, the problem is that when I try to divide them the array is actually divided, except for when the "-" sign is present, in fact if I have as a string "1 + 1 * 1 -1"
the result will be an array ["1", "1", "1-1"]
while it should be ["1", "1", "1", "1"]
Thanks everyone in advance.
ANSWER
Answered 2021-Jun-14 at 14:05QUESTION
I have a mathematical expression given as a String
and I have to extract all the variables which are identified as a letter, possibly followed by a number (e.g x
or x0
). It works for simple expressions but if I try it with a more complicated equation I pick also numbers which I don't want since my goal is to determinate if the two equations use the same variables.
ANSWER
Answered 2021-Jun-14 at 22:05The expression keeps the digits because they are not included in the regex search for the split
method when creating the String variable
.
Try splitting at one or many non-alphanumeric characters (\W+
), which may be followed by zero or many digits (\d*
).
"\W+\d*"
Adding \d*
to the end of your existing regex should also work.
"[^a-z0-9?]\d*"
Tested on regex101 with Java 8.
Please let me know whether this resolved your question.
QUESTION
I have a CSS-type cubic bezier where
A = 0,0
B = a, b where 0<1
C = p, q where 0
I want a function pointOnBezier(a,b,p,q,x) => {x:x, y:y}
I've found this which helps but no mathematical formula😓
https://upload.wikimedia.org/wikipedia/commons/d/db/B%C3%A9zier_3_big.gif
Any help would be appreciated.
Thanks in advance!!!
ANSWER
Answered 2021-Jun-15 at 09:12You can find the Cubic Bézier curve formula in Wikipedia.
Once you have the formula, you need to find t by x and then find y by t. To find t, you need to solve a cubic equation. You can find the code for solving cubic equation from other places such as this post.
Here is the code for your reference:
QUESTION
I would like to include the mathematical sign for a multiplication (like the lower-case x) in a plot title.
I have tried using latex2exp
to no avail - can someone help me with this?
The reprex code is this:
...ANSWER
Answered 2021-Jun-14 at 16:15One approach might be to use the unicode code for the multiplication symbol:
QUESTION
I have two dataframes, one looks like this
...ANSWER
Answered 2021-Jun-14 at 12:19You can explode the map column and group by key to sum up the lengths:
QUESTION
Problem:
I'm looking for a catch-all function that I can use to calculate the tangent of any polynomial function at x. I'm indifferent to the language used although JavaScript or Python would be prefered! I should be able to pass in any x value and an array of coefficients in the format, a + bx + cx^2 + dx^3 ... and so on.
Example function format:
...ANSWER
Answered 2021-Jan-30 at 00:21Okay so after a day of struggling with it I think I have got the solution in both JavaScript and Python!
QUESTION
I am implementing a words/2
predicate in which a list of characters can be rendered to a list of the character as a word inside a list. I use the mathematical symbol <=>
to denote that they're working in any mode. Please advise if there's a better expression.
The example:
...ANSWER
Answered 2021-Jun-07 at 14:32split(_, [], [[]]).
split(C, [C|Xs], [[]|Ys]) :-
split(C, Xs, Ys).
split(C, [X|Xs], [[X|Y]|Ys]) :-
split(C, Xs, [Y|Ys]).
QUESTION
I need to compute tf.Variable
gradients in a class method, but use those gradients to update the variables at a later time, in a different method. I can do this when not using the @tf.function
decorator, but I get the TypeError: An op outside of the function building code is being passed a "Graph" tensor
error when using @tf.function
. I've searched for understanding on this error and how to resolve it, but have come up short.
Just FYI if you're curious, I want to do this because I have variables that are in numerous different equations. Rather than trying to create a single equation that relates all the variables, it is easier (less computationally costly) to keep them separate, compute the gradients at a moment in time for each of those equations, and then incrementally apply the updates. I recognize that these two approaches are not mathematically identical.
Here is my code (a minimal example), followed by the results and error message. Note that when gradients are computed and used to update variables in a single method, .iterate()
, there is no error.
ANSWER
Answered 2021-Jun-11 at 18:28Please check the quick fix below corresponding to your question.
QUESTION
I need to declare an array of variables to be used later, but without initializing any of the variables inside, is it possible?. I need this to make a function use easier as i have to perform some mathematical operations on the vector varlist, which is the input of another function. Can anybody help?
...ANSWER
Answered 2021-Jun-11 at 11:02I think it's better to use the index of the list as variable : variable[index]
Or if you want to use named variables... use a Dictionary{key : value} instead of list.
QUESTION
I have a mathematical problem: these five strings are IDs for the same object. Due to these differences, objects appear multiple times in my Access table/query. Although there are a lot of these mutations, but I take this as an example.
...
ANSWER
Answered 2021-Jun-11 at 05:12As @Vincent has suggested, look at using a custom function to convert all of the different data to be consistent. Based on what you have described, the following seems to work:
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