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I am reading the paper "Transformation Invariance in Pattern Recognition – Tangent Distance and Tangent Propagation", where the tangent vector is calculated for the given curve $s(P,\alpha)$ at $\alpha=0$ by differentiating with respect to $\alpha$, that is, $\frac{\partial s(P,\alpha)}{\partial\alpha}$. For the curve, I have taken one $2D$ image and I am rotating it with matrix $R=\left[\matrix{cos(\alpha)\space -sin(\alpha)\\sin(\alpha) \space\space\space\space\space cos(\alpha)} \right]$.

As my image is fixed, the curve is just a function of $\alpha$. Therefore, to find the tangent vector, what I am doing is as follows:

  1. I am rotating the image by the matrix $R^{'}$ which is $R^{'}=\left[\matrix{-sin(\alpha)\space -cos(\alpha)\\cos(\alpha) \space\space\space\space\space -sin(\alpha)} \right]$

  2. This rotates the image by $90$ degree, which is not the expected result.

I have done the same exercise by differentiating numerically and I am getting the expected answer which is as follows:
Picture <span class=$\alpha$=0">
Tangent vector plot at <span class=$\alpha$=0">

Please, help me to understand my mistake in taking derivative of the matrix and multiplying it with image.

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  • $\begingroup$ What does the rotation have to do with the differentiation of the curve function? $\endgroup$ – nbro Feb 27 at 14:31
  • $\begingroup$ You can see 2D image as a point in 16 X 16 dimensional space. So if you keep rotating you get a curve in 256 dimensional space wrt rotation angle. So I want to differentiate this curve wrt to rotational angle. $\endgroup$ – Arun Chauhan Feb 28 at 3:55

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