Step 1

Determining the best regression model by using finite differences. The method of finite differences is used for curve fitting with polynomial models. It is best introduced by an example as shown in table given below. For the sequence of x value inputs at the regularly spaced points \(\displaystyle-{3},-{2},-{1},\cdots\ {3},\) the y value outputs are shown in the table.

Table 1 is given below.

From the table, we can understand that this is third difference finite function of third degree.

This explains

\(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}.\)

Step 2

Here we need to find the value of a, b, c and d

\(\displaystyle{a}={\frac{{-{6}}}{{{3}!}}}\)

\(\displaystyle{a}={\frac{{-{6}}}{{{6}}}}\)

\(\displaystyle{a}=-{1}\)

Step 3

Taking \(\displaystyle{x}={0},-{1},{1}\) for calculating the value of rest of the co-efficients.

\(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}\)

\(\displaystyle{f{{\left(-{1}\right)}}}={1}+{b}-{c}+{6}\)

\(\displaystyle{x}=-{1}\)

\(\displaystyle{y}={0}\)

\(\displaystyle{0}={b}-{c}+{7}\)

\(\displaystyle-{7}={\left({b}-{c}\right)}\)

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{y}&\text{First Difference}&\text{Second Difference}&\text{Third Difference}\backslash{h}{l}\in{e}-{3}&{0}&\text{EMPTY VALUE}&\text{EMPTY VALUE}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}-{2}&-{4}&-{4}-{0}=-{4}&\text{EMPTY VALUE}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}-{1}&{0}&{0}-{\left(-{4}\right)}={4}&{4}-{\left(-{4}\right)}={8}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}{0}&{6}&{6}-{0}={6}&{2}&{2}-{8}=-{6}\backslash{h}{l}\in{e}{1}&{8}&{8}-{6}={2}&-{4}&-{6}\backslash{h}{l}\in{e}{2}&{0}&{0}-{8}=-{8}&-{10}&-{6}\backslash{h}{l}\in{e}{3}&-{24}&-{24}-{0}={24}&-{16}&-{6}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

Also

\(\displaystyle{f{{\left({1}\right)}}}=\ -{1}+{b}+{c}+{6}\)

\(\displaystyle{8}={b}+{c}+{5}\)

\(\displaystyle{b}+{c}={3}\)

Moreover,

\(\displaystyle{f{{\left({0}\right)}}}=-{0}+{0}+{0}+{d}\)

\(\displaystyle{x}={0}\)

\(\displaystyle{y}={6}\)

\(\displaystyle{6}={d}\)

Thus, finding b and c by subtracting and adding equation (1) and (2)

\(\displaystyle{2}{b}=-{4}\)

\(\displaystyle{b}=-{2}\)

\(\displaystyle{2}{c}={10}\)

\(\displaystyle{c}={5}\)

At last the finite function becomes:

\(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}-{2}{x}^{{{2}}}+{5}{x}+{6}\)

Step 4

Sketching a grph from the function found above

Thus, end behaviour of \(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}-{2}{x}^{{{2}}}+{5}{x}+{6}\)

is described from the table as below

As \(\displaystyle\rightarrow\infty,\ {f{{\left({x}\right)}}}\rightarrow\ -\infty\) and As \(\displaystyle{x}\rightarrow\ -\infty,{f{{\left({x}\right)}}}\rightarrow\infty\)

Determining the best regression model by using finite differences. The method of finite differences is used for curve fitting with polynomial models. It is best introduced by an example as shown in table given below. For the sequence of x value inputs at the regularly spaced points \(\displaystyle-{3},-{2},-{1},\cdots\ {3},\) the y value outputs are shown in the table.

Table 1 is given below.

From the table, we can understand that this is third difference finite function of third degree.

This explains

\(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}.\)

Step 2

Here we need to find the value of a, b, c and d

\(\displaystyle{a}={\frac{{-{6}}}{{{3}!}}}\)

\(\displaystyle{a}={\frac{{-{6}}}{{{6}}}}\)

\(\displaystyle{a}=-{1}\)

Step 3

Taking \(\displaystyle{x}={0},-{1},{1}\) for calculating the value of rest of the co-efficients.

\(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}+{b}{x}^{{{2}}}+{c}{x}+{d}\)

\(\displaystyle{f{{\left(-{1}\right)}}}={1}+{b}-{c}+{6}\)

\(\displaystyle{x}=-{1}\)

\(\displaystyle{y}={0}\)

\(\displaystyle{0}={b}-{c}+{7}\)

\(\displaystyle-{7}={\left({b}-{c}\right)}\)

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{y}&\text{First Difference}&\text{Second Difference}&\text{Third Difference}\backslash{h}{l}\in{e}-{3}&{0}&\text{EMPTY VALUE}&\text{EMPTY VALUE}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}-{2}&-{4}&-{4}-{0}=-{4}&\text{EMPTY VALUE}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}-{1}&{0}&{0}-{\left(-{4}\right)}={4}&{4}-{\left(-{4}\right)}={8}&\text{EMPTY VALUE}\backslash{h}{l}\in{e}{0}&{6}&{6}-{0}={6}&{2}&{2}-{8}=-{6}\backslash{h}{l}\in{e}{1}&{8}&{8}-{6}={2}&-{4}&-{6}\backslash{h}{l}\in{e}{2}&{0}&{0}-{8}=-{8}&-{10}&-{6}\backslash{h}{l}\in{e}{3}&-{24}&-{24}-{0}={24}&-{16}&-{6}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

Also

\(\displaystyle{f{{\left({1}\right)}}}=\ -{1}+{b}+{c}+{6}\)

\(\displaystyle{8}={b}+{c}+{5}\)

\(\displaystyle{b}+{c}={3}\)

Moreover,

\(\displaystyle{f{{\left({0}\right)}}}=-{0}+{0}+{0}+{d}\)

\(\displaystyle{x}={0}\)

\(\displaystyle{y}={6}\)

\(\displaystyle{6}={d}\)

Thus, finding b and c by subtracting and adding equation (1) and (2)

\(\displaystyle{2}{b}=-{4}\)

\(\displaystyle{b}=-{2}\)

\(\displaystyle{2}{c}={10}\)

\(\displaystyle{c}={5}\)

At last the finite function becomes:

\(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}-{2}{x}^{{{2}}}+{5}{x}+{6}\)

Step 4

Sketching a grph from the function found above

Thus, end behaviour of \(\displaystyle{f{{\left({x}\right)}}}=-{x}^{{{3}}}-{2}{x}^{{{2}}}+{5}{x}+{6}\)

is described from the table as below

As \(\displaystyle\rightarrow\infty,\ {f{{\left({x}\right)}}}\rightarrow\ -\infty\) and As \(\displaystyle{x}\rightarrow\ -\infty,{f{{\left({x}\right)}}}\rightarrow\infty\)