Understanding Taylor Series Notation: A Clear Explanation

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Hello everyone. I Very Excite to be here : D!

Ok, so I was reading on taylor polynomials for several variables and came across the following on wikipedia:

<math>T(\mathbf{x}) = f(\mathbf{a}) + \mathrm{D} f(\mathbf{a})^T (\mathbf{x} - \mathbf{a}) + \frac{1}{2!} (\mathbf{x} - \mathbf{a})^T \mathrm{D}^2 f(\mathbf{a}) (\mathbf{x} - \mathbf{a}) + \cdots\!
</math>

Or, please follow the link:
http://en.wikipedia.org/wiki/Taylor_series#Taylor_series_in_several_variables

*it is the second to last expression in that section.

What do they mean by that "T" on top of the matrices? what does it tell us to do?

Thanks.
 
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frasifrasi said:
What do they mean by that "T" on top of the matrices? what does it tell us to do?

Thanks.

I'm looking at the Wiki article. Usually, that symbol is used to denote the transpose of a matrix. Does that work?
 
Oh, I see. Thanks.
 
In this case, the superscript T is a bit superfluous, and maybe even incorrect. As the gradient of some vector lies in the dual of the vector space, the gradient of a column vector is typically represented as a row vector (and vice versa). Using that convention, it is better to say \nabla \mathbf f(a) \mathbf f(a) (i.e., no transpose). Even better, use Einstein sum notation.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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