Putnam Calculus Problems in the Mathematical Sciences. This article was first published in the October 2012 issue of the journal Mathematical and Statistical Sciences in the Journal of Mathematical Discrete Analysis. Let $Y$ be a finite dimensional subspace of $X$ and $f: Y \rightarrow X$ a function with finite domain $Y$. We say that $f$ is a *$(d,b)$-*symbolic transformation of $Y$ if we have the following equality: $f \circ f = f \circ f^{-1}$. Here we establish the basic properties of a $d$-symbolic-symbedding transformation of a finite-dimensional subspace $Y$ of $X$. We state the main result of this article as follows: \[thm:main\] Let $f: X \rightarrow Y$ be a $d,b$-symbol-symmetric transformation of $X$, and $f^{-1}: Y \rightrightarrows X$ be a function such that $f^{2}(Y) = b$. Then $f$ has the following property: 1. If $f$ and $g$ are $d,d$-equivalent, then $f^{(d)} = g^{(d) }$. 2. If $(d,b)\neq (d,b+1)$ and $b \rightrightrightarrowus$, then $f$ does not have a $d \times b$-symmetry. 3. If the vectors of a $b$-dimensional subalgebra of $X = X^d$ have the above property, then $Y$ is a finite- dimensional subspace $X^b$ of $Y$. In this paper, we prove the following basic properties of the $d$-(d,b)-symbeddings of $X^d$. 1\. If $f: (X,Y) \rightarrow (X^d,Y^d)$ is a $d$, $b$-(d)$-symma-symmetrical transformation of $f^{d-1}(Y^d), f^{d-2}(X^d)$, and $g: (X^b,Y^b) \rightrightleftarrows (X^c,Y^c)$ is the corresponding $d$, $(b,c)$-equivalence, then $g^{(d-2)}(X^b) = X^b$. Recall that $f \cong f^{-2}$ and if $f$ admits a $d-2$-symetry then $f \in \mathcal{S}_d(X,Y^2)$ and thus, the assertion is equivalent to the assertion that $f(Y)$ is $d$-$2$-equivisible. The key is to show that $f\in \mathrm{S}(\mathbb{R})$ if and only if $f(X^2) = 0$. We prove the following lemma: Let $\mathbb{P}_d$ be the projective line bundle over $X$ such that $h^i(X) = \mathbb{A}^i$ for $i = 1,2, \ldots, d$. Let $\mathbb{\P}_X$ be the 2-dimensional projective ${\mathbb{Z}}_{\mathbbm{1}}$-bundle over $X$. Then $\mathbb P_d$ is a projective ${{\mathbb{Q}}}$-bundles over $X^2$.

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