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Koushik Vemula
MathWorks
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I am Applicant Support Engineer at MathWorks. My responsibility is to provide the best support for an application like ML, Computer Vision, Image Processing, and Signal Processing.
Disclaimer: Any articles/ideas/opinions here are my own and in now way reflect that of MathWorks.
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Optimtool not enough arguments
According to my understanding you are facing this problem because of the way you are trying to access the function. You need...
presque 5 ans il y a | 0
Can I derive a dependent variable in Matlab?
According to my understanding you want to create a variable ‘y’ which is dependent on variable ‘x’ You can do it using ‘syms’...
presque 5 ans il y a | 1
| A accepté
cumulative maximum loss code
According to my understanding you have a variable ‘x’ which has ‘n’(say) number of values. You would like to find the value of...
presque 5 ans il y a | 0
Make probability matrix from data set
According to my understanding, you’d like to have a matrix (say ‘Prob’) which stores the data values in an n-by-3 matrix where f...
presque 5 ans il y a | 0
how to access compass formatting details?
In MATLAB there is a "polarplot" function that allows you to update the properties of the polar axes. Please see the following d...
presque 5 ans il y a | 1
How can I convert this pseudo code to matlab?
According to what I understand the pseudo code you’ve written is correct with some minor points to be taken under consideration....
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Manipulating matrix with 'for' loop
According to what I understand you want to change the values of 'main diagonal’ and ‘adjacent diagonals’ of the given matrix. ...
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Find n for inverse of e approximation
According to what I understand, you are trying to get the value of ‘accurate_n’ which gives you a tolerance of ‘0.0001’. The pro...
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How to take a derivative of x with respect to time.
According to what I understand, you are trying to get the differentiation of variable ‘o’ with respect to the parameter ‘t’. ...
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Why does x(x+y) dy give a x^3 component?
Both answers are correct. d/dy[ x*(x+y)^2/2 ] = d/dy[ (x^3)/2 + x^2*y + y^2*x/2 ] = x*(x + y) d/dy[ x*y*(2*x+y))/2 ] = x*(x ...
presque 5 ans il y a | 3
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