Let $f$ be some function in some programming language (like C), and we need $n$ bits to store this function. Suppose we have some fixed value $v$ for the argument, then let g() { f(v) }be the function with no argument which just calles $f$ with $v$ as an argument. Then we need $n + |v| + C$ bits to store it, where $C$ is some constant which takes account of the additional bits needed to declare $g$, and this is independent of $v$ and $f$.So in any programming language so far, to go from programs accepting parameters, to the one for a fixed parameter the desriptional complexity  raises up at most by a fixed constant plus the descriptional length of $v$.Does the same hold for Turing-machines? If I have a Turing-Machine which for input $x$ produces $y$, then to have a machine without accepting parameter to produce $y$ I can built one that first writes $x$ and then runs $M$, and a general machine that writes $x$ for any $x$ would be one that saves the tokens of $x$ in its states and writes them one after another. If I look at how instructions are specified in Turing machines, then we see that for $x$ we need $|x|\cdot C$ bits (with $C > 1$) then to store this additional machine, hence given $M$ with $M(x) = y$ we get a machine $M_x = y$ without arguments with $$ |M| + C|x| + C'$$needed bits, where $C > 1$ and $C'$ account for some additional setup independent of $M$ and $x$.But would it be possible to built a Turing machine $M_x$ with $M_x = y$ without input which has descriptional complexity at most $|M| + |x| + C'$ for some constant $C'$ independent of $M$ and $x$? Like above for ordinary programming languages?
For a turing machine which computes $y$ with argument $x$, how much does the descriptional length of a machine producing $y$ without input increases
formal languages;computability;turing machines;kolmogorov complexity;descriptive complexity