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Looking at the definition of tangent cone in continuous optimization :

If $M$ is a open subset of $\mathbb R^n$ $x \in M$, The tangent cone of $M$ at $x$ is defined by $$\mathbb T (M, x) = \big\{d \in \mathbb R^n | \exists x_k \subset M , \eta_k \subset \mathbb R : \eta_k \to 0 , x_k \to x \text{ and } \frac{x_k-x}{\eta_k} \to d\big\} $$

I don't understand the definition , what is actually happening here ? Can anyone explain me elaborately .

I kind of make sense if i consider $M=\{y : y=x^2\} \subset \mathbb R^2$ . But for a general set $M$ i don't get the idea.

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One thing to observe is that $\mathbb T(M,x)$ indeed has the property of an (abstract) cone: if $d\in \mathbb T(M,x)$, then every positive multiple of $d$ is also in $\mathbb T(M,x)$. For example, $2d$ is in the tangent cone because you can use the same $x_k$ that gave you $d$, and replace $\eta_k$ with $\frac12\eta_k$.

(By the way, the definition you have does not preclude $\eta_k$ from being negative. I think this is a mistake.)

So, to understand what vectors $\mathbb T(M,x)$ contains, it is enough to understand what unit vectors it contains. Then the rest of them are the multiples of those unit vectors.

The reason to focus on the unit vectors in $\mathbb T(M,x)$ is that they are easier to find: we can take $\eta_k = \|x_k-x\|$ in the definition. In other words, a unit vector $u$ belongs to $\mathbb T(M,x)$ if and only if there is a sequence $x_k\to x$ such that $$\frac{x_k-x}{\|x_k-x\|} \to u \tag{1}$$ Try to visualize (1): you stand at $x$ looking at approaching $x_k$, and mark the directions from which they hit you. The set of all these directions (normalized vectors $x_k-x$) gives you all the unit vectors contained in the tangent cone.

For example, let $ M$ be the sphere $u^2+v^2+w^2=4$. What is its tangent cone at the North Pole $(0,0,2)$? We must consider points of the sphere approaching the North pole. They can be written as $(u_k,v_k,\sqrt{4-u_k^2-v_k^2})$. The difference vector is $$(u_k,v_k,\sqrt{4-u_k^2-v_k^2} - 2) = \left(u_k,v_k, \frac{ -u_k^2-v_k^2}{\sqrt{4-u_k^2-v_k^2}+ 2}\right) $$ The length of this vector is approximately $\sqrt{u_k^2+v_k^2}$, up to the error $O(u_k^2+v_k^2)$ (the third coordinate is negligibly small). So, when the vector is normalized, the result is close to $(u_k, v_k,0)/\sqrt{u_k^2+v_k^2} $. This shows that the tangent cone is the plane $w=0$.

A less smooth example (the surface of the circular cone) is described here.

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This was a very nice answer. I'm sorry it was not accepted for some reason! – mathmath8128 May 25 '15 at 4:06

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