trendline option (excel) in matlab 
Here is a working solution:
a = [15.5156,0.1995;
7.6003,0.2999;
9.4829,0.2592;
12.2185,0.2239;
23.4094,0.1811];
x = a(:, 1);
y = a(:, 2);
n = 2; % order of the fitted polynomial trendline
p = polyfit(x, y, n);
m = 1000; % number of trendline points (the larger the smoother)
xx = linspace(min(x), max(x), m);
yy = polyval(p, xx);
figure;
hold on;
scatter(a(:,1), a(:,2));
plot(xx, yy, 'r');
You can easily put the trendline calculator code into a separate function.

Solving a simple nonlinear equation in Matlab 
I assume that x is a symbolic variable, while the vector z is numeric? So,
something like this:
syms x y;
z = 1:5;
solve(exp(x*max(z))/sum(exp(x*z))==y,x)
which returns
Warning: The solutions are parametrized by the symbols:
z1 = RootOf(y*z^4  z^4 + y*z^3 + y*z^2 + y*z + y, z)
> In solve at 180
ans =
log(z1)
solve does give you a solution, just not an explicit one. It tells you that
you have several solutions that are the roots of some polynomial. The more
elements that z has (and depending on the specific values of z) the higher
the order the polynomial will be. The roots of quadratics and cubics are
easily found. Quartics (with constant coefficients) can also be factored.
Some quintics can be too, but for higher orders, there are very few known
analytical solutions for

how to find nonlinear equation parameters using a dataset 
I think the correct format should be:
Y(X) = a [ b + (c * X^d)]^e
A nice calculator is Wolfram Alpha, very detailed.

Templated linear equation system solver, C++ 
Take a look into Eigen:
"Eigen is a C++ template library for linear algebra: matrices, vectors,
numerical solvers, and related algorithms"
http://eigen.tuxfamily.org/index.php?title=Main_Page

Solving a simple sparse linear system of equation using csparse: cs_cholsol 
Having never worked with csparse before, I skimmed the source code.
When you call cs_spalloc() to create Operator, you are creating a triplet
(indicated by setting the last parameter to 1). But, after the call to
cs_copmress(), the result is no longer a triplet (you can detect this by
checking the result and see that Operator>n is now 1 after
compression). So, it is an error to traverse the matrix as if it were.
You can use the cs_print() API to print your sparse matrix.
As an aside, your code leaks memory, since the compressed matrix is a new
allocation, and the original uncompressed matrix was not freed by
cs_compress().

Gradient descent does not return incorrect prediction for linear function 
In this line
sum = sum + ((dot(theta', x(i, :))  y(j)) * x(i,j));
you are using a wrong index of y, it should be y(i), as j is a dimension
iterator, not the sample iterator.
After the change
theta =
1.5168e07
1.0000e+00

Excel VBA Modify Existing Equation 
if the formula is changed to =A1+A3+A5 then code could be:
Range("B1").Formula = Range("B1").Formula & "+" &
Range("A7").Address
or, without dollar signs,
Range("B1").Formula = Range("B1").Formula & "+" &
Range("A7").Address(False, False)
Added in response to OPs comment. If I assume the new cell is two rows
above where the formula is, which is the currently the active cell:
ActiveCell.Formula = Replace(ActiveCell.Formula, ")", "," & _
ActiveCell.Offset(2, 0).Address(False, False) & ")")
Adjust the value 2 as necessary.

Create a trendline via flot? 
Here is a trend line plugin for flot, but it may need some additional work:
flot issue 270 (download link after the first post)

How to apply power fit trendline in R or gnuplot? 
To give an example for curve fitting with gnuplot, consider the following
data set data.txt:
0.5 0.8
2.4 9.3
3.2 37.9
4.9 68.2
6.5 155
7.8 198
The fit with a power law function may look like this:
set termoption enhanced
f(x) = a*x**b;
fit f(x) 'data.txt' via a,b
plot 'data.txt' with points title 'data points',
f(x) with lines title sprintf('power fit curve f(x) = %.2f·x^{%.2f}',
a, b)
With the terminal settings
set terminal pngcairo size 800,600 font ',12'
this gives the result
This is, of course, the most basic way to fit, the 'specialties' depend on
your actual needs.

Incorrect date formating in Excel worksheet 
You may use this snippit to convert from excel serial date format
protected static DateTime FromExcelSerialDate(int serialDate)
{
if (serialDate > 59) serialDate = 1; //Excel/Lotus 2/29/1900
bug
return new DateTime(1899, 12, 31).AddDays(serialDate);
}

Solving linear and nonlinear regression for vectors 
Found a good solution: DLIB solves this problem quite well. There is an
example here:
http://dlib.net/least_squares_ex.cpp.html
Note: if you need to use matrices, DLIB still requires data to be stored as
vectors. That is, if you are solving Ab=c and A is a 3x3 matrix,
parameter_vector in the example linked above has to be typedef
matrix<double,9,1> parameter_vector; or it won't compile.

different result in webkitlineargradient and mozlineargradient 
This generator should produce you the correct crossbrowser code.
You can use import from css button at the lower right corner of the
generated css fieldset. For your code, this was the output:
background: #696969; /* Old browsers */
background: mozlineargradient(top, #696969 0%, #3d3d3d 50%, #292929
50%, #171717 100%); /* FF3.6+ */
background: webkitgradient(linear, left top, left bottom,
colorstop(0%,#696969), colorstop(50%,#3d3d3d), colorstop(50%,#292929),
colorstop(100%,#171717)); /* Chrome,Safari4+ */
background: webkitlineargradient(top, #696969 0%,#3d3d3d 50%,#292929
50%,#171717 100%); /* Chrome10+,Safari5.1+ */
background: olineargradient(top, #696969 0%,#3d3d3d 50%,#292929
50%,#171717 100%); /* Opera 11.10+ */
background: mslineargradient(top, #696969 0%,#3

Excel file created in Excel 2007 is not opening in Excel 2003 using c# 
If you make an Excel file programmatically and name it with the .xls
extension from Excel 2007 or 2010, you may be making an Open XML (.xlsx)
file unknowingly, as this is the default save format for those versions of
Excel. Essentially, if you are hard coding the file name, you may be
saving an .xlsx file with the .xls extension, which would explain why you
can't open the files. Try renaming the files generated from Excel 2007 to
the .xlsx extension and see if you can open them.

Display Excel inside VB6 / VB.NET Form & work with Excel without excel menu 
As far as I can tell  this is not possible. The reason being that you
cannot subclass the excel window in VBA.
If you program outside of excel with VB.NET or VB6 it would be sort of
possible, but would probably not work really good.
If you only want to display some data from a sheet and perform some small
actions on them, you could probably read the Data via the EPPLUS Library,
display them as a datatable (EPPLUS can convert between datatable and excel
files), perform your actions and save them to the file afterwards. But this
would only work in VB.net

Why do I keep getting zero in my equation? 
You might want to try
int percent = (100*rateNum) / count;
or use floating point for the math. Your value is being rounded down to
zero and then never recovering.

How this does equation simplify to 
As you have written it, I think the equation is wrong. Since * represents
component wise multiplication:
(rs, gs, bs) * (as, as, as) = as(rs, gs, bs)
(rd, gd, bd) * (1as, 1as, 1as) = (1as)(rd, gd, bd)
So:
(rs, gs, bs) * (as, as, as) + (rd, gd, bd) * (1as, 1as, 1as)
= as(rs, gs, bs) + (1as)(rd, gd, bd)

If equation has no roots 
Is it a program or a function? If it's a program, something that will be
invoked by people, the right way is to output the phrase "No solutions" or
something like that.
Now, if it's a function that returns variable(s), the question is
different. First, not all languages have a None as a possible numeric
value; C/C++, for example, does not. Does the code solve any kind of
equation? In that case, consider this. An equation may have multiple roots.
That means you should somehow return a collection of roots. If there are no
roots, an empty collection would be the right thing to return.
Also, an equation may have an infinite number of roots (example: 0*x=0).

PHP simple equation 
Change your if statement to:
// security
if($prv != 0 AND $prv != 2) {
die("<p>Error</p>");
}
With the OR it will always evaluate as true because when $prv == 0 it does
not equal 2. Makes sense?

Using a value from a matrix as a multiplier in an equation? 
In MATLAB, apostrophes ('') define a string. If the name of your matrix is
answer, you can refer to its second value by the command answer(2) as
mentioned by @Schorsch. For more infos on vectors and matrices, you can
check this site.

Quadratic Equation program 
Lines of this form are suspect;
if ( sqrt((b*b)  4*a*c) > 0 )
as you will be calling sqrt on a negative.
You should check the real case, for example, by doing this:
if (b*b  4*a*c > 0)
I've also dropped the unnecessary brackets. * has higher precedence than .
Only call the sqrt function on a nonnegative parameter.

How can I implement this equation in python? 
Before implementing your own statistical stuff, have you looked into SciPy?
SciPy Stats has many modules implemented using NumPy (C written, so it's
faster than Python,
but callable from Python, with a Pythonic syntax). So maybe you don't need
to write any code yourself.

Solve equation with a set of points 
Since you have more points than unknown coefficients, you'll want to do a
least squares fit. This calculation will give you the coefficients that
minimize the mean square error between the function and all the points.
So substitute your 50 points into the assumed equation and get 50 equations
for 3 coefficients. This is easily expressed as a matrix:
Ax = b
where the unknown vector x are your coefficients.
Premultiply both sides by the transpose of A and solve using a matrix
solver.
Start by plotting the data you have. The choice of equation you start with
will make your job easier or harder. Are you certain about that leading
term? If that wasn't appropriate, you could take the natural log of both
sides and it's a simple linear equation. The leading term makes that
harder to te

Collision equation for two 3d objects 
Referencing http://en.wikipedia.org/wiki/Elastic_collision :
for i in range (0, len(allObjects) ):
mass1 = getVolume(allObjects[i],objMultiplySize)
velocity1 =
getVelocity(allObjects[i],originalTime,objMultiplySize)
mass2 = getVolume(objSel[j],objMultiplySize)
velocity2 =
getVelocity(objSel[j],originalTime,objMultiplySize)
v1new = velocity1; //just initialization
v2new = velocity2; //just initialization
v1new[0] = (velocity1[0] *( mass1mass2) +
2*mass2*velocity2[0])/(mass1 + mass2);
v2new[0] = (velocity2[0] *( mass2mass1) +
2*mass1*velocity1[0])/(mass1 + mass2);
v1new[1] = (velocity1[1] *( mass1mass2) +
2*mass2*velocity2[1])/(mass1 + mass2);

Difference Equation Blowing Up  C 
I notice that this line:
k = (float)(i/(segmentsl1));
is performing integer division. Since i<segmentsl it will be the case
that i/segmentsl is 0. This is probably not what you intend.
It is also the case that farther in your program, you use
tempvstime[i][j+1] and tempvstime[i][j], while both should be j+1.
A modified version of your program which runs correctly and has altered
variable names and fewer parentheses in the math follows.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(void){
int width, maxt;
float rvalue;
int i,t;
float k;
//Pull in initial data
printf("Enter as integers without spaces: Number of Segments  Length,
Number of Segments  time ,value
");
scanf("%d,%d,%f", &width,&maxt,&rvalu

Add an equation into scatter plots in R 
The "ps"object is being handled as an expression, i.e. not being
evaluated. To get around this use bquote and .()
legend(locator(1), legend= bquote("Linear model: "* R^2*.(ps)),
text.col= "red",cex=1,bty="n")
BTW the first version would be more compactly represented without the
paste:
legend(locator(1), expression(Linear~model*":"~ R^2 == 0.1234),
text.col= "black",cex=1,bty="n")
The only thing needing to be quoted is the semicolon.

Showing mathjax equation one by one 
It seems working, but solution is not elegant.
typesetMath(jQuery(".formula"),0);
function typesetMath(formulas,ind){
MathJax.Hub.Queue(["Typeset",MathJax.Hub,formulas.get(ind)],
function(){
jQuery(formulas.get(ind)).css("visibility","visible");
if(ind+1<formulas.length)
typesetMath(formulas,ind+1);
});
}

Matplotlib Image fit to equation? 
Since you are saving a figure that represents a string, it might be better
to remove the black frame box and make the background transparent, which is
done by adding these two lines,
plt.figure(frameon=False)
plt.axes(frameon=0)
To make the figure size fit the equation, save the figure this way,
plt.savefig('D:/latex.png', bbox_inches='tight')
Finally, its better to remove the figure from memory after saving it, and
this is done by adding this line,
plt.close()
So the new code would be,
import matplotlib.pyplot as plt
def convert(string):
plt.figure(frameon=False)
plt.axes(frameon=0)
if string[0] != '$' and string[1] != '$':
string = '$' + string + '$'
plt.text(0.01, 0.8, string, fontsize=50)
plt.xticks(())
plt.yticks(())
plt.savefig('D:/late

Solving equation in Python 3.3 
Y, C, m, and d are all strings. You want to convert them into ints first:
Y = int(f[2:4])
C = int(f[:2])
...
Are you sure that equation even works, though? It looks like it'd produce a
lot of noninteger weekdays. You might've miscopied it. Also, brackets
aren't a grouping operator in Python; they're the list construction syntax.
You'll want to replace those brackets with parentheses in the expression
for w. (Or were those brackets supposed to be the floor operator? If so,
you'll want math.floor, from the math module, or just int if truncation is
fine.)

Best way to solve an equation in form f(x)=K*u(x) 
Your equations are not yet in standard form, yet are reducible to it.
Compute
>> syms u1 u2 f3 f4;
>> K = magic(4);
>> f = [2 3 f3 f4].';
>> u = [u1 u2 9 7].';
>> K*u  f
ans =
16*u1 + 2*u2 + 116
5*u1 + 11*u2 + 143
9*u1 + 7*u2  f3 + 138
4*u1 + 14*u2  f4 + 142
In which I altered the formatting for clarity. From this we can see that if
we define
b = [116 143 138 142].';
A = [16 2 0 0
5 11 0 0
9 7 1 0
4 14 0 1];
Then we have an equation in standard form
A*X = b
where
X = [u1 u2 f3 f4].'
Then, it is a simple matter of
>> X = A
ans =
5.963855421686747e+000 % u1
1.028915662650603e+001 % u2
+1.230120481927711e+00

SELECT *, sum() with equation only returns one row 
In MySQL you can include columns in the select that are not aggregated and
not in the group by. Because your query has an aggregation function, MySQL
recognizes it as an aggregation query, and only returns one row.
Actually, you need to bring in the total value for a calculation on each
row. Here is one way using a subquery in the select statement:
SELECT *,
(select sum(q6) as sumq6 from table where field2 = 'xxx') / (`q1` *
(`q1` + `q2` + `q3` + `q4` + `q5` + `q6`) / 6)
* 100) AS percent
FROM table
WHERE field2 = 'xxx'
ORDER BY `percent` ASC;
Here is another way, using a cross join:
SELECT *,
(sumq6 / (`q1` * (`q1` + `q2` + `q3` + `q4` + `q5` + `q6`) / 6)
FROM table cross join
(select sum(q6) as sumq6 from table where field2 = 'xxx') as const
WHERE field2 = 'xxx

How can I get this array equation to work? 
I assume your reuse of i is tripping up your loop. At the end of the loop,
when for handles the i, it may have completely changed value.
If var i = (+document.getElementById('if' + i).value)/12; sets i to some
value greater than 24, the loop will terminate at the start of the next
iteration due to the i <= 24 condition.
Instead, use a different variable name for your loop index and the values
in your computation.

Need help porting this equation to java 
Your problem is inside Math.pow  you're multiplying your index value by a
scalar each time.
4 * Math.pow(i, 3)
And
3 * Math.pow(i, 2)
Should fix it.
Edit: And the integer division mentioned in other answers.

+/ equation with a single textfield 
Because, You are performing addition with an array type variable.
$assoc = mysql_fetch_assoc($sql);
Here, $assoc is an array variable so try like this,
$v_nu = $v_stk + $assoc['v_stk'];

Why has changed both sides of equation in java? 
d=dataset
You now have two variables that both refer to the same object.
Unlike C++, Java never implicitly copies anything.
If you want a separate copy of an object, you need to make one yourself.
If the object implements Cloneable, you can do that using clone().

How can I solve this kind of equation in Maple? 
In (Eq. 1) it's not your syntax that's an issue. You have three unknowns
{a,b,c} but only one equation. You simply do not have enough equations to
determine {a,b,c} uniquely. Maple's solve function only returns an answer
(if possible) if the number of variables equals the number of equations.
In (Eq. 2) you use square brackets, which are used for ordered lists. The
solve function requires a set of equations, which are indicated by curly
braces. Again, you have three variables but only one equation. Same
problem.
If the equations are linear (which they aren't in your case), Maple can
find a parameterization for the solutions in the case of an underdetermined
system:
http://www.maplesoft.com/support/help/Maple/view.aspx?path=solve/linear.

Custom equation solver error 
Here, the map which you are passing is not with the generic types. Hence,
get() will always return an object, which is not an appropriate argument
for parseInt() method.
Changing the method signature to
public boolean solve(String equation, Map< String ,String > gladMap) {
should solve the errors.

Write mathematical equation in flash 
You should consider using LaTeX2SWF library instead of trying to import
Microsoft Equation directly.
Example here
If you can't download it, you can try something like this
You should also convert all of your Math Equations from Microsoft format to
LaTeX. There are a lot of automated programs and plugins, like this, but
there are lots of others

How to fit equation of the form y=ao+a1logx+a2log(2/x) 
What toolboxes are available to you?
The easiest way would probably be the cftool. (Type it into your command
window) if you have the curve fitting toolbox. But polyfit should do as
well.
The main problem I see: Your coefficients are not independent of one
another. Because log(2/x) is equal to log(2)  log(x) your equation
becomes:
y = ao + a1*log(x) + a2*log(2)  a2*log(x);
which is equivalent to:
y = bo + b1*log(x);
Try that one.

Wrong answer with equation in python 
Integer division. 4/3 evaluates to 1 as it is rounded down.
Use 4.0 instead to force floating point arithmetic:
>>> 4.0/3 * 6.67*1e11*3.14*6378000*5515
9.822550047279998
or use Python 3, where floating point division is the default, or use from
__future__ import division to achieve the same in Python 2:
Python 2.7.5 (default, May 22 2013, 12:00:45)
[GCC 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2336.11.00)] on
darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from __future__ import division
>>> 4/3 * 6.67*1e11*3.14*6378000*5515
9.822550047279998
This behaviour is documented under the Binary arithmetic operators section:
The / (division) and // (floor division) operators yield the quotient of
their arguments.

Program for the discriminant of a quadratic equation 
One problem I can see is with the following 
return new complexNumber[]{ (b + Math.sqrt(discriminant))/(2*a)};
the result of (b + Math.sqrt(discriminant))/(2*a) is not a complexNumber.
An array of complexNumber should have instance of complexNumber.
