tensor algebra $T(\mathbb{R})$ I would like to understand the product on the tensor algebra $T(V)$ by a concrete example. I know the general construction and the product is defined just by juxtaposition of the elements.  
So if I set $V = \mathbb{R}$ as a real vector space, then 
$$T(\mathbb{R}) = \bigoplus_{i\geq 0} \mathbb{R}$$
since $V^{\otimes i} \cong \mathbb{R}$
My question is, how do the elements of $T(\mathbb{R})$ look like and how is the algebra multiplication expressed in this particular case.
 A: You can identify $\mathcal T(\mathbb R)$ with the Algebra of Polynomials $\mathbb R[x]$. As stated in the answer of @Fallen_Apart, an element in $ \mathcal T(\mathbb R)$ is just a finite sum of real number which have assigned a degree $i$ (indicating in which of the spaces $\mathcal T^i(\mathbb R)\cong\mathbb R$ they sit. Now you can just use these as the coefficients of a polynomial and identify $\mathcal T(\mathbb R)$ with $\mathbb R[x]$ in that way. (Alternatively, any linear map $f$ from $\mathbb R$ to an associative algebra $A$ over $\mathbb R$ with unit defines a unique homomorphism $\mathbb R[x]\to A$ mapping $x$ to $f(1)$.)
However, $\mathcal T(\mathbb R)$ is rather misleading as an example for a tensor algebra since it is commutative because $\mathbb R$ has dimension $1$. If start from $\mathbb R^2$, then you can view the tensor algebra as "non-commutative polynomials" in two variables. 
A: 
My question is, how do the elements of $T(\mathbb{R})$ look like

You just said that $T(\mathbb{R}) = \bigoplus_{i\geq 0} \mathbb{R},$ hence any element of $T(\mathbb{R})$ is of form $r_{i_1}+\dots +r_{i_k}$ where $r_{i_j}\in V^{\otimes i_j}\cong\mathbb{R}.$ Here $+$ signs are taken from direct product $T(\mathbb{R}) = \bigoplus_{i\geq 0} \mathbb{R}.$ 

and how is the algebra multiplication expressed in this particular case.

Since $T(\mathbb{R})$ forms graded algebra ($\otimes:T^k(\mathbb{R})\otimes T^l(\mathbb{R})\rightarrow T^{k+l}(\mathbb{R})$) it is sufficient to tell how it behaves at arbitrary degree. If $r_k\in T^k(\mathbb{R})$ and $r_l\in T^l(\mathbb{R}),$ then $r_k\otimes r_l=r_kr_l.$ (This equality is due to the fact we mutually identified $\mathbb{R}\otimes\mathbb{R}\cong\mathbb{R}$ by isomorphism $r\otimes s\mapsto rs$)
To be even more evident lests compute something more complex. Consider $r_1,r_3,r_4,r_5,s_1,s_2,s_5$ ($r_i,s_i\in T^i(\mathbb{R})$) and compute $(r_1+r_3+r_4+r_5)\otimes(s_1+s_2+s_5).$ It equals
$$r_1s_1+r_1s_2+r_3s_1+(r_3s_2+r_4s_1)+(r_1s_5+r_4s_2)+r_3s_5+r_4s_5.$$
Note that $+$ sings inside brackets are additions of real numbers, when other $+$ signs means $+$ taken from direct product $T(\mathbb{R}) = \bigoplus_{i\geq 0} \mathbb{R}$ in the same way as in the beginnig of an answer.
