Recall that the mixed-type quadratic and cubic functional equation (1.9) was introduced in [41] and the authors proved the following theorem.

Theorem 2.1

Let X and Y be vector spaces. A function (h:Xlongrightarrow Y) satisfies functional equation (1.9) if and only if there exist a quadratic function (A^{2}:Xlongrightarrow Y), a cubic function (A^{3}:Xlongrightarrow Y) and a constant (A^{0}) such that

$$ h(x)= A^{0}+A^{2}(x)+A^{3}(x) $$

for all (xin X).

Next, by using an alternative method we obtain the general solution of (1.9), which is a tool to reach some of our goals in this section.

Lemma 2.2

Let X and Y be real vector spaces. Suppose that (h:Xlongrightarrow Y) satisfies (1.9) for all (x,yin X).

  1. (i)

    If h is even and (h(0)=0), then it is quadratic;

  2. (ii)

    If h is odd, then it is cubic.


(i) We first note that the evenness of h converts (1.9) to

$$ h(x+2y)-3h(x+y)+3h(x)-h(x-y)=0, $$


for all (x,yin X) (here and in the rest of the proof). Letting (x=0) in (2.1), we have

Interchanging x with (x-y) in (2.1), we find

$$ h(x+y)-3h(x)+3h(x-y)-h(x-2y)=0. $$


A difference computation of (2.1) and (2.3) shows that

$$ h(x+2y)+h(x-2y)-4h(x+y)-4h(x-y)+6h(x)=0. $$


Replacing x by 2x in (2.4) and using (2.2), we obtain

$$ h(x+y)+h(x-y)-h(2x+y)-h(2x-y)+6h(x)=0. $$


Switching ((x,y)) by ((y,x)) in (2.5) and applying again the evenness of f, we obtain

$$ h(x+y)+h(x-y)-h(x+2y)-h(x-2y)+6h(y)=0. $$


Inserting (2.4) into (2.6), we find

$$ h(x+y)+h(x-y)=2h(x)+2h(y). $$

(ii) Using our assumption on (1.9), we obtain

$$ h(x+2y)-3h(x+y)+3h(x)-h(x-y)-6h(y)=0. $$


Putting (x=0) in (2.7) and applying from the oddness of h, we find

On the other hand, if we replace (x-y) instead of x in (2.6), then,

$$ h(x+y)-3h(x)+3h(x-y)-h(x-2y)-6h(y)=0. $$


Now, it follows from (2.7) and (2.9) that

$$ h(x+2y)+h(x-2y)=4bigl[h(x+y)+h(x-y)bigr]-6h(x). $$

This completes the proof. □

Henceforth, let V and W be vector spaces over (mathbb{Q}), (nin mathbb{N}) and (x_{i}^{n}=(x_{i1},x_{i2},ldots ,x_{in})in V^{n}), where (iin {1,2}). We denote (x_{i}^{n}) by (x_{i}) if there is no risk of a mistake. Let (x_{1},x_{2}in V^{n}) and (min mathbb{N}_{0}) with (0leq mleq n). Put (mathcal{M}^{n}= { mathfrak{N}_{n}=(N_{1},ldots ,N_{n})mid N_{j} in {x_{1j}pm x_{2j},x_{1j}} } ), where (jin {1,ldots ,n}). Consider

$$ mathcal{M}_{m}^{n} := bigl{ mathfrak{N}_{n} in mathcal{M}^{n}mid operatorname{Card}{N_{j}: N_{j}=x_{1j}}=m bigr} . $$

For (rin mathbb{Q}), we put (rmathcal{M}_{m}^{n}= { rmathfrak{N}_{n}: mathfrak{N}_{n}in mathcal{M}_{m}^{n} } ) in which (rmathfrak{N}_{n}=(rN_{1},ldots ,rN_{n})).

Definition 2.3

A mapping (f:V^{n}longrightarrow W) is n-multicubic or multicubic if f satisfies (1.3) in each variable.

For a multicubic mapping f, we use the notations

$$ f bigl(mathcal{M}_{m}^{n} bigr):= sum_{mathfrak{N}_{n}in mathcal{M}_{m}^{n}}f(mathfrak{N}_{n}), $$



$$ f bigl(mathcal{M}_{m}^{n},z bigr):=sum _{mathfrak{N}_{n}in mathcal{M}_{m}^{n}}f(mathfrak{N}_{n},z)quad (zin V). $$

We recall that (binom{n}{m}:=frac{n!}{m!(n-m)!}) is the binomial coefficient, where (n, min mathbb{N}_{0}) with (ngeq m). In the upcoming result, we find a necessary and sufficient condition for a several-variable mapping to be multicubic.

Proposition 2.4

For a mapping (f: V^{n}longrightarrow W), the following assertions are equivalent.

  1. (i)

    f is multicubic;

  2. (ii)

    f satisfies

    $$ sum_{tin {-2,2}^{n}} f(x_{1}+tx_{2})= sum_{m=0}^{n}4^{n-m}(-6)^{m}f bigl( mathcal{M}_{m}^{n}bigr), $$


    where (f (mathcal{M}_{m}^{n} )) is defined in (2.10).


(mbox{(i)}Rightarrow mbox{(ii)}) The proof of this implication is by induction on n. For (n=1), it is clear that f fulfills (1.3). Suppose that (2.11) holds for some positive integer (n>1). Then,

$$begin{aligned} &sum_{tin {-2,2}^{n+1}} fbigl(x_{1}^{n+1}+tx_{2}^{n+1} bigr) \ &quad =4sum_{tin {-2,2}^{n}}sum _{sin {-1,1}} fbigl(x_{1}^{n}+tx_{2}^{n},x_{1,n+1}+sx_{2,n+1} bigr)-6 sum_{tin {-2,2}^{n}} fbigl(x_{1}^{n}+tx_{2}^{n},x_{1,n+1} bigr) \ &quad =4sum_{m=0}^{n}sum _{sin {-1,1}}4^{n-m}(-6)^{m}fbigl( mathcal{M}_{m}^{n},x_{1,n+1}+sx_{2,n+1} bigr)-6 sum_{m=0}^{n}4^{n-m}(-6)^{m}f bigl(mathcal{M}_{m}^{n},x_{1,n+1}bigr) \ &quad =sum_{m=0}^{n+1}4^{n+1-m}(-6)^{m}f bigl(mathcal{M}_{m}^{n+1}bigr). end{aligned}$$

(mbox{(ii)}Rightarrow mbox{(i)}) Let (jin {1,ldots ,n}) be arbitrary and fixed. It is enough to prove that f is cubic in the jth variable. Set

$$ f^{*}_{j}(z):=f (z_{1},ldots ,z_{j-1},z,z_{j+1},ldots ,z_{n} ). $$

Assuming (x_{2m}=0) for all (min {1,ldots ,n}backslash {j}), (x_{2j}=w) and (x_{1}= (z_{1},ldots ,z_{j-1},z,z_{j+1},ldots ,z_{n} )) in (2.11), we obtain

$$begin{aligned} &2^{n-1} bigl[f^{*}_{j}(z+2w)+f^{*}_{j}(z-2w) bigr] \ &quad =sum_{m=0}^{n-1}left [begin{pmatrix} n-1 \ m end{pmatrix}2^{n-1-m}4^{n-m} (-6)^{m}right ] bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) bigr] \ &qquad {}+sum_{m=1}^{n}left [ begin{pmatrix} n \ m end{pmatrix}2^{n-m}4^{n-m} (-6)^{m} right ]f^{*}_{j}(z) \ &quad =4sum_{m=0}^{n-1}left [begin{pmatrix} n-1 \ m end{pmatrix}8^{n-1-m}(-6)^{m}right ] bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) bigr] \ &qquad {}-6sum_{m=0}^{n-1}left [begin{pmatrix} n \ m end{pmatrix}8^{n-1-m}(-6)^{m}right ]f^{*}_{j}(z) \ &quad =4times 2^{n-1} bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) bigr]-6 times 2^{n-1}f^{*}_{j}(z). end{aligned}$$


It follows from (2.12) that

$$ f^{*}_{j}(z+2w)+f^{*}_{j}(z-2w)=4 bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) bigr]-6f^{*}_{j}(z). $$

Now, the proof is completed. □

Definition 2.5

Let (nin mathbb{N}) and (kin {0,ldots ,n}). A mapping (f:V^{n}longrightarrow W) is called k-quadratic and (n-k)-cubic (briefly, multiquadratic–cubic) if f is quadratic (see equation (1.2)) in each of some k variables and is cubic in each of the other variables (see equation (1.3)).

In Definition 2.5, we suppose for simplicity that f is quadratic in each of the first k variables, but one can obtain analogous results without this assumption. It is obvious that for (k=n) (resp., (k=0)), the above definition leads to the so-called multiquadratic (resp., multicubic) mappings; some basic facts on the mentioned mappings can be found, for instance, in [8, 13] and [45].

To reach our results in the rest of this section, we identify (x=(x_{1},ldots ,x_{n})in V^{n}) with ((x^{k},x^{n-k})in V^{k}times V^{n-k}), where (x^{k}:=(x_{1},ldots ,x_{k})) and (x^{n-k}:=(x_{k+1},ldots ,x_{n})). Put (x_{i}^{k}=(x_{i1},ldots ,x_{ik})in V^{k}) and (x_{i}^{n-k}=(x_{i,k+1},ldots ,x_{in})in V^{n-k}), where (iin {1,2}). For a multiquadratic–cubic mapping f, we also recall the notation

$$ f bigl(x_{i}^{k},mathcal{M}_{m}^{n-k} bigr):=sum_{mathfrak{N}_{n} in mathcal{M}_{m}^{n-k}}f bigl(x_{i}^{k}, mathfrak{N}_{n-k} bigr), $$


$$ mathcal{M}_{m}^{n-k} := bigl{ mathfrak{N}_{n-k} in mathcal{M}^{n-k}mid operatorname{Card}{N_{j}: N_{j}=x_{1j}}=m bigr} , $$

in which

$$ mathcal{M}^{n-k}= bigl{ mathfrak{N}_{n-k}=(N_{k+1}, ldots ,N_{n})mid N_{j}in {x_{1j}pm x_{2j},x_{1j}} bigr} . $$

In the following result, we describe a multiquadratic–cubic mapping as an equation. The proof is similar to the proof of [8, Proposition 2.1], but we include some parts for the sake of completeness.

Proposition 2.6

Let (nin mathbb{N}) and (kin {0,ldots ,n}). If a mapping (f: V^{n}longrightarrow W) is kquadratic and (n-k)cubic mapping, then f satisfies the equation

$$ sum_{sin {-1,1}^{k}}sum _{tin {-2,2}^{n-k}} f bigl(x_{1}^{k}+sx_{2}^{k},x_{1}^{n-k}+tx_{2}^{n-k} bigr)=2^{k}sum_{m=0}^{n-k}4^{n-k-m}(-6)^{m} sum_{iin {1,2}}f bigl(x_{i}^{k}, mathcal{M}_{m}^{n-k} bigr) $$


for all (x_{i}^{k}=(x_{i1},ldots ,x_{ik})in V^{k}) and (x_{i}^{n-k}=(x_{i,k+1},ldots ,x_{in})in V^{n-k}), where (iin {1,2}).


Since for (kin {0,n}) our assertion follows from Proposition 2.4 and [45, Theorem 3], we can assume that (kin {1,ldots ,n-1}). Let (x^{n-k}in V^{n-k}) be arbitrary and fixed. Consider the mapping (g_{x^{n-k}}:V^{k}longrightarrow W) defined via (g_{x^{n-k}} (x^{k} ):=f (x^{k},x^{n-k} )) for (x^{k}in V^{k}). Similar to the proof of Proposition 2.1 from [8], one can show that

$$ sum_{sin {-1,1}^{k}}f bigl(x_{1}^{k}+sx_{2}^{k},x^{n-k} bigr)=2^{k} sum_{j_{1},ldots ,j_{k}in {1,2}}f bigl(x_{j_{1}1},ldots ,x_{j_{k}k},x^{n-k} bigr) $$


for all (x_{1}^{k},x_{2}^{k}in V^{k}) and (x^{n-k}in V^{n-k}). Similar to the above, we obtain from Proposition 2.4 that

$$ sum_{tin {-2,2}^{n-k}} f bigl(x^{k},x_{1}^{n-k}+tx_{2}^{n-k} bigr)=sum_{m=0}^{n-k}4^{n-k-m}(-6)^{m}f bigl(x^{k},mathcal{M}_{m}^{n-k} bigr) $$


for all (x_{1}^{n-k},x_{2}^{n-k}in V^{n-k}) and (x^{k}in V^{k}). Now, equalities (2.14) and (2.15) show that (2.13) holds for f. □

It is easily seen that the mapping (f:mathbb{R}^{n}longrightarrow mathbb{R}) defined by (f(r_{1},ldots ,r_{n})=prod_{j=1}^{k}prod_{i=k+1}^{n}r_{j}^{2}r_{i}^{3}) is multiquadratic–cubic and hence (2.13) is valid for f by Proposition 2.6. Therefore, this equation is called a multiquadratic–cubic functional equation. Note that in the case (k=n) and (k=0), equation (2.13) converts to (1.7) and (2.11), respectively.

Definition 2.7

Let V and W be vector spaces over (mathbb{Q}), (nin mathbb{N}). A mapping (f:V^{n}longrightarrow W) is said to be n-multimixed quadratic–cubic, or briefly multimixed quadratic–cubic, if f satisfies (1.9) in each variable.

Let (x_{1},x_{2}in V^{n}) and (p_{l}in mathbb{N}_{0}) with (0leq p_{l}leq n), where (lin {1,2,3,4}). Set

$$ mathbb{M}^{n}= bigl{ mathfrak{M}_{n}=(M_{1}, ldots ,M_{n})mid M_{j} in {x_{1j}pm x_{2j},x_{1j},x_{2j},-x_{2j}} bigr} , $$

for all (jin {1,ldots ,n}). Consider the subset (mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}) of (mathbb{M}^{n}) as follows:

$$begin{aligned} mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} : =&bigl{ mathfrak{M}_{n} in mathbb{M}^{n}mid operatorname{Card}{M_{j}: M_{j}=x_{1j}}=p_{1}, \ & operatorname{Card}{M_{j}: M_{j}=x_{2j} }=p_{2}, operatorname{Card}{M_{j}: M_{j}=-x_{2j} }=p_{3}, \ & operatorname{Card}{M_{j}: M_{j}=x_{1j}+x_{2j} }=p_{4}bigr} . end{aligned}$$

Hereafter, for a multimixed quadratic–cubic mapping f, we use the notations

$$ f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} bigr):= sum_{ mathfrak{M}_{n}in mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}}f ( mathfrak{M}_{n} ), $$



$$ f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},z bigr):=sum _{ mathfrak{M}_{n}in mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}}f ( mathfrak{M}_{n},z )quad (zin V). $$

For each (x_{1},x_{2}in V^{n}), we consider the equation

$$ f(x_{1}+2x_{2})=sum _{p_{1}=0}^{n}sum_{p_{2}=0}^{n-p_{1}} sum_{p_{3}=0}^{n-p_{1}-p_{2}} sum _{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} bigr), $$


where (f (mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} )) is defined in (2.16).

Definition 2.8

Given a mapping (f: V^{n}longrightarrow W). We say f

  1. (i)

    has zero condition if (f(x)=0) for any (xin V^{n}) with at least one component that is equal to zero;

  2. (ii)

    is even in the jth variable if

    $$ f(z_{1},ldots ,z_{j-1},-z_{j},z_{j+1}, ldots , z_{n})=f(z_{1}, ldots ,z_{j-1},z_{j},z_{j+1}, ldots , z_{n}); $$

  3. (iii)

    is odd in the jth variable if

    $$ f(z_{1},ldots ,z_{j-1},-z_{j},z_{j+1}, ldots , z_{n})=-f(z_{1}, ldots ,z_{j-1},z_{j},z_{j+1}, ldots , z_{n}). $$

In what follows, it is assumed that every mapping (f: V^{n}longrightarrow W) satisfying (2.17) has zero condition. With this assumption, we unify the general system of mixed-type quadratic and cubic functional equations defining a multimixed quadratic–cubic mapping to an equation and indeed this functional equation describe a multimixed quadratic–cubic mapping as well.

Proposition 2.9

A mapping (f: V^{n}longrightarrow W) is multimixed quadratic–cubic if and only if it satisfies equation (2.17).


Suppose that f is a multimixed quadratic–cubic mapping. We proceed with the proof by induction on n. For (n=1), it is obvious that f satisfies equation (1.9). Let (2.17) be true for some fixed and positive integer (n>1). Then,

$$begin{aligned} &f bigl(x_{1}^{n+1}+2x_{2}^{n+1},z bigr) \ &quad =sum_{p_{1}=0}^{n} sum _{p_{2}=0}^{n-p_{1}} sum_{p_{3}=0}^{n-p_{1}-p_{2}} sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},z bigr), end{aligned}$$


for all (x_{1},x_{2}in V^{n}) and (zin V). Using (2.18) and the fact that (2.17) holds for the case (n=1), we obtain

$$begin{aligned} &f bigl(x_{1}^{n+1}+2x_{2}^{n+1} bigr) \ &quad =f bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1}+2x_{2,n+1} bigr) \ &quad =3f bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1}+x_{2,n+1} bigr)+f bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1}-x_{2,n+1} bigr) \ &qquad {} -3f bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1} bigr)+3f bigl(x_{1}^{n}+2x_{2}^{n},x_{2,n+1} bigr)-3f bigl(x_{1}^{n}+2x_{2}^{n},-x_{2,n+1} bigr) \ &quad =3sum_{p_{1}=0}^{n} sum _{p_{2}=0}^{n-p_{1}} sum _{p_{3}=0}^{n-p_{1}-p_{2}} sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{1,n+1}+x_{2,n+1} bigr) \ &qquad {} +sum_{p_{1}=0}^{n} sum _{p_{2}=0}^{n-p_{1}} sum _{p_{3}=0}^{n-p_{1}-p_{2}} sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{1,n+1}-x_{2,n+1} bigr) \ &qquad {} -3sum_{p_{1}=0}^{n} sum _{p_{2}=0}^{n-p_{1}} sum _{p_{3}=0}^{n-p_{1}-p_{2}} sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{1,n+1} bigr) \ &qquad {} +3sum_{p_{1}=0}^{n} sum _{p_{2}=0}^{n-p_{1}} sum _{p_{3}=0}^{n-p_{1}-p_{2}} sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{2,n+1} bigr) \ &qquad {} -3sum_{p_{1}=0}^{n} sum _{p_{2}=0}^{n-p_{1}} sum _{p_{3}=0}^{n-p_{1}-p_{2}} sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},-x_{2,n+1} bigr) \ &quad =sum_{p_{1}=0}^{n+1} sum _{p_{2}=0}^{n+1-p_{1}} sum _{p_{3}=0}^{n+1-p_{1}-p_{2}} sum_{p_{4}=0}^{n+1-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n+1} bigr). end{aligned}$$

This means that (2.17) holds for (n+1).

Conversely, let (jin {1,ldots ,n}) be arbitrary and fixed. Set

$$ f^{*}_{j}(z):=f (z_{1},ldots ,z_{j-1},z,z_{j+1},ldots ,z_{n} ). $$

Putting (x_{2k}=0) for all (kin {1,ldots ,n}backslash {j}), (x_{2j}=w) and (x_{1}= (z_{1},ldots ,z_{j-1},z,z_{j+1},ldots ,z_{n} )) in (2.17), we obtain

$$begin{aligned} f^{*}_{j}(z+2w)&=left [sum_{p_{1}=0}^{n-1}sum _{p_{4}=1}^{n-p_{1}} begin{pmatrix} n-1 \ p_{1} end{pmatrix} begin{pmatrix} n-1-p_{1} \ p_{4}-1 end{pmatrix}(-3)^{p_{1}}3^{p_{4}}right ]f^{*}_{j}(z+w) \ &quad {} +left [sum_{p_{1}=0}^{n-1} sum_{p_{4}=0}^{n-1-p_{1}}begin{pmatrix} n-1 \ p_{1} end{pmatrix} begin{pmatrix} n-1-p_{1} \ p_{4} end{pmatrix}(-3)^{p_{1}}3^{p_{4}}right ]f^{*}_{j}(z-w) \ &quad {}+left [sum_{p_{1}=1}^{n} sum_{p_{4}=0}^{n-p_{1}}begin{pmatrix} n-1 \ p_{1}-1 end{pmatrix} begin{pmatrix} n-p_{1} \ p_{4} end{pmatrix}(-3)^{p_{1}}3^{p_{4}}right ]f^{*}_{j}(z) \ &quad {}+3left [sum_{p_{1}=0}^{n-1} sum_{p_{4}=0}^{n-1-p_{1}}begin{pmatrix} n-1 \ p_{1} end{pmatrix} begin{pmatrix} n-1-p_{1} \ p_{4} end{pmatrix}(-3)^{p_{1}}3^{p_{4}}right ]f^{*}_{j}(w) \ &quad {}-3left [sum_{p_{1}=0}^{n-1}sum _{p_{4}=0}^{n-1-p_{1}}begin{pmatrix} n-1 \ p_{1} end{pmatrix} begin{pmatrix} n-1-p_{1} \ p_{4} end{pmatrix}(-3)^{p_{1}}3^{p_{4}}right ]f^{*}_{j}(-w). end{aligned}$$


On the other hand,

$$begin{aligned} &sum_{p_{1}=0}^{n-1} begin{pmatrix} n-1 \ p_{1} end{pmatrix}(-3)^{p_{1}}sum_{p_{4}=0}^{n-1-p_{1}} begin{pmatrix} n-1-p_{1} \ p_{4} end{pmatrix}3^{p_{4} +1}times 1^{n-1-p_{4}} \ &quad =3sum_{p_{1}=0}^{n-1} begin{pmatrix} n-1 \ p_{1} end{pmatrix}(-3)^{p_{1}}4^{n-1-p_{1}}=3(4-3)^{n-1}=3. end{aligned}$$



$$ sum_{p_{1}=0}^{n-1} begin{pmatrix} n-1 \ p_{1} end{pmatrix}(-3)^{p_{1}}sum_{p_{4}=0}^{n-1-p_{1}} begin{pmatrix} n-1-p_{1} \ p_{4} end{pmatrix}3^{p_{4}}times 1^{n-1-p_{4}}=1. $$



$$ sum_{p_{1}=0}^{n-1} begin{pmatrix} n-1 \ p_{1} end{pmatrix}(-3)^{p_{1}+1}sum_{p_{4}=0}^{n-1-p_{1}} begin{pmatrix} n-1-p_{1} \ p_{4} end{pmatrix}3^{p_{4}}times 1^{n-1-p_{4}}=-3. $$


It follows from (2.19), (2.20), (2.21) and (2.22) that

$$ f^{*}_{j}(z+2w)=3f^{*}_{j}(z+w)+f^{*}_{j}(z-w)-3f^{*}_{j}(z)+3f^{*}_{j}(w)-3f^{*}_{j}(-w). $$

This completes the proof. □

Corollary 2.10

Suppose that a mapping (f: V^{n}longrightarrow W) satisfies equation (2.17).

  1. (i)

    If f is even in each variable, then it is multiquadratic. Moreover, f satisfies equation (1.7);

  2. (ii)

    If f is odd in each variable, then it is multicubic. In addition, equation (2.11) is true for f;

  3. (iii)

    If f is even in each of some k variables and is odd in each of the other variables, then it is multiquadratic–cubic. In particular, f fulfilling equation (2.13).


(i) It is shown in Proposition 2.9 that for each j, (f^{*}_{j}) satisfies (1.9). Since it is assumed that (f^{*}_{j}(0)=0), the result follows from part (i) of Lemma 2.2.

(ii) This is a direct consequence of part (ii) of Lemma 2.2.

(iii) The result follows from the previous parts. □

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit


This article is autogenerated using RSS feeds and has not been created or edited by OA JF.

Click here for Source link (