simple math question

4 ビュー (過去 30 日間)
LIU WEIHUA
LIU WEIHUA 2011 年 11 月 18 日
There has a function can be written as: y=H*[X,Y,1]'. where [X,Y,1]is known constants. I want finding the derivatives of the series of matrix H=A*[r1,r2,t]; where A is a 3 by 3 matrix,and r1,r2,t is 3 by 1 column . During the derivatives process, A ,r1,r2,t should be viewed as the variables separately. Is anyone know how to tackle it with matlab? Or just give me some tips on this.Thank you very much!
  2 件のコメント
Andrew Newell
Andrew Newell 2011 年 11 月 18 日
You want partial derivatives with respect to the matrix A? That doesn't make sense.
LIU WEIHUA
LIU WEIHUA 2011 年 11 月 20 日
sorry ,I express wrong ,it's derivatives instead of partial derivatives.

サインインしてコメントする。

回答 (1 件)

Alex
Alex 2011 年 11 月 20 日
If I understand correctly, you have H as a function of r1, r2, & t.
By this, you have 3 partial derivatives, dH/dr1, dH/dr2, & dH/dt.
If those 3 partial derivative are what you want, then this is done much easier on paper.
let, A be a matrix with elements aij, with i being the row and j be the column.
(in this case, I am assuming the matrix A has only constants, folling the same steps will get you the results for variables with higher powers).
Then, H is a system of 3 equations with 3 unknowns. Looking at the first row,
a11*r1 + a12*r2 + a13*t = H1,
then dH/dr1 becomes d(a11*r1) = dH1/dr1
then dH/dr2 becomes d(a12*r2) = dH1/dr2
then dH/dt becomes d(a13*t) = dH1/dt
(the other 2 rows follow the same).
These 9 results can be combined back into a dH matrix, with each row representing one of the partial derivatives.
  1 件のコメント
Alex
Alex 2011 年 11 月 20 日
This can be explained more in depth at: http://en.wikipedia.org/wiki/Matrix_calculus

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeMathematics についてさらに検索

タグ

タグが未入力です。

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by