Understanding Derivative of $\Vert f\Vert_{2}^{2}$

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In summary, the given conversation discusses the derivative of the p-norm squared evaluated at p=2. The process involves using the chain rule and plugging in p=2 at the end, resulting in the final expression of 1/2 times the integral of the absolute value of f squared multiplied by the natural logarithm of the ratio of f squared to the 2-norm squared of f.
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[tex]$\frac{d}{dp}(\Vert f\Vert_{p}^{2})\mid_{p=2}=\frac{d}{dp}((\int_{\mathbb{\mathbb{R}}^{n}}\mid f\mid^{p}dx)^{\frac{2}{p}})\ldots=\frac{1}{2}\int_{\mathbb{\mathbb{R}}^{n}}\mid f(x)\mid^{2}ln\left(\frac{\mid f(x)\mid^{2}}{\Vert f\Vert_{2}^{2}}\right)$[/tex]

this is the derivative evaluated at p=2.

does anyone see how this works? I started with the e^ln(whole integral) and then chain rule and so on...then plugged in p=2 at the end but couldn't get it.wow i cannot get the d/dp to show up properly something is wrong with the latex...
 
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cheeez said:
[tex]\frac{d}{dp}(\Vert f\Vert_{p}^{2})\mid_{p=2}=\frac{d}{dp}((\int_{\mathbb{\mathbb{R}}^{n}}\mid f\mid^{p}dx)^{\frac{2}{p}})\ldots=\frac{1}{2}\int_{\mathbb{\mathbb{R}}^{n}}\mid f(x)\mid^{2}ln\left(\frac{\mid f(x)\mid^{2}}{\Vert f\Vert_{2}^{2}}\right)$[/tex]

this is the derivative evaluated at p=2.

does anyone see how this works? I started with the e^ln(whole integral) and then chain rule and so on...then plugged in p=2 at the end but couldn't get it.





wow i cannot get the d/dp to show up properly something is wrong with the latex...

Did that fix it?
 

FAQ: Understanding Derivative of $\Vert f\Vert_{2}^{2}$

What is the meaning of "derivative of $\Vert f\Vert_{2}^{2}$"?

The derivative of $\Vert f\Vert_{2}^{2}$ is a mathematical concept that represents the rate of change of the squared Euclidean norm of a function $f$ in a given direction. It measures how much the value of $\Vert f\Vert_{2}^{2}$ changes as the input variable changes.

Why is the derivative of $\Vert f\Vert_{2}^{2}$ important?

The derivative of $\Vert f\Vert_{2}^{2}$ is important because it allows us to optimize functions by finding the values of the input variables that minimize or maximize the function. It also helps us understand the behavior of a function and make predictions about its future values.

How do you calculate the derivative of $\Vert f\Vert_{2}^{2}$?

The derivative of $\Vert f\Vert_{2}^{2}$ can be calculated using the chain rule, which states that the derivative of a composite function is equal to the product of the derivative of the outer function and the derivative of the inner function. In this case, the inner function is $f$ and the outer function is the squared Euclidean norm.

What is the relationship between the derivative of $\Vert f\Vert_{2}^{2}$ and the gradient of $f$?

The derivative of $\Vert f\Vert_{2}^{2}$ is closely related to the gradient of $f$, which is a vector that points in the direction of the greatest increase of $f$. The gradient of $f$ is equal to twice the derivative of $\Vert f\Vert_{2}^{2}$, and its direction is the same as the direction of the derivative.

Can the derivative of $\Vert f\Vert_{2}^{2}$ be negative?

Yes, the derivative of $\Vert f\Vert_{2}^{2}$ can be negative. This means that as the input variable increases, the value of $\Vert f\Vert_{2}^{2}$ decreases. In other words, the function $f$ is decreasing in that direction. A negative derivative can also indicate a local maximum of the function.

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