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$P_n(x,y)$ denote the vector space of polynomial with degree less than equal to $n $, then $\dim (P_n(x,y))=\dfrac{(n+1)(n+2)}{2!}$

How this answer is coming?

My attempt,

Total base elements (of degree${}=n)= n+1$

Total base elements (of degree${}=n-1)= n$

$\ldots$

Total base elements (of degree${}=1)= 2$.

Therefore total possible base elements${}=\dfrac{((n+1)+1)(n+1)}{2}$

Am I Right?

How to generalize the formula. Please explain.

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In general, the space of polynomials in $k$ variables having degree $\le n$ has the basis:

$$\{x_1 ^{a_1} \cdots x_k^{a_k}: \sum_{i=1}^k a_i \le n\}$$

which is the same as the set:

$$\{1^{a_0} x_1^{a_1} \cdots x_k ^{a_k}: \sum_{i=0}^k a_i = n\}$$

Using stars and bars, the cardinality of this set (hence the dimensionality of the vector space) is:

$${n + k \choose n}$$

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  • $\begingroup$ Can you give hints, how this generalised formula is coming?(proof) $\endgroup$ – Unknown x Aug 1 '17 at 4:06
  • $\begingroup$ @Maneesh.Maths it is a straightforward application of the stars and bars theorem. Make sure to understand the theorem well, then the formula will be very clear. $\endgroup$ – Cauchy Aug 1 '17 at 4:07
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The univariate polynomial ring $k[x]$ has dimension $n+1$ with basis $\{1,x,\dots,x_n\}$.

For each one of these basis vectors $x^i$, we can multiply by $y^j$ with $j \leq n-i$.

For the first one $x^0$, this is $n$ vectors (any power of $y$), then $x^1$, we can mulyiply $n-1$ vectors, etc.

From this, we get that the dimension is $\sum_{k=0}^{n+1} n-k=\sum_{k=0}^{n+1} k=(n+1) \cdot (n+2)/2$.

here is a generalization using stars and bars-- the basic idea is you just want to find out how many different monomials you can have so that the sum of their exponents ((total degree) is less than $n$

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Hint you have to find the numbers of $x^iy^j$, $i+j\leq n$, for $i=0$ you have $n+1$ choices for $j$,... for $i=n$ you have one choice, so in total you have:

$$0+1+\cdots+n+1={{(n+1)(n+2)}\over 2}.$$

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$$ \begin{array}{cccccccc} & a_{0,0} & + & a_{0,1} x & + & a_{0,2} x^2 & + & a_{0,3} x^3 \\ & & & & & & \swarrow \\ + & a_{1,0} y & + & a_{1,1} xy & + & a_{1,2} xy^2 \\ & & & & \swarrow\\ + & a_{2,0} y^2 & + & a_{2,1}xy^2 \\ & & \swarrow \\ + & a_{3,0} y^2 \end{array} $$ The arrows link terms of degree $3.$

Another diagonal sequence of arrows, parallel to this one, would link terms of degree $2.$

And yet another of terms of degree $1,$ just two of those in this case.

So the number of coefficients is the "triangular number" $1+2+3+4,$ and $4$ is just $1$ more than the degree.

In other words, your reasoning is correct.

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