Q27E
Question
Show that the set of functions\(\left\{ {{\bf{1,}}\,{\bf{x,}}\,{{\bf{x}}^{\bf{2}}}{\bf{,}}...{\bf{,}}\,{{\bf{x}}^{\bf{n}}}} \right\}\), where n is a positive integer, is linearly independent on every open interval (a, b).
[Hint: Use the fact that a polynomial of degree at most n has no more than n zeros unless it is identically zero.]
Step-by-Step Solution
VerifiedThus, this set forms the linearly independent set on every open interval (a, b).
A polynomial of degree n,
\({{\bf{a}}_{\bf{0}}}{\bf{1 + }}{{\bf{a}}_{\bf{1}}}{\bf{x + }}{{\bf{a}}_{\bf{2}}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}...{\bf{ + }}{{\bf{a}}_{\bf{n}}}{{\bf{x}}^{\bf{n}}}{\bf{ + }}...{\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {\bf{1}} \right)\)
Substitute \({\bf{x = 0}}\)in the above expression,
\({{\bf{a}}_{\bf{0}}}{\bf{ = 0}}\)
From equation (1),
\({{\bf{a}}_{\bf{1}}}{\bf{x + }}{{\bf{a}}_{\bf{2}}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}...{\bf{ + }}{{\bf{a}}_{\bf{n}}}{{\bf{x}}^{\bf{n}}}{\bf{ + }}...{\bf{ = 0}}\)
Take the derivative with respect to x,
\({{\bf{a}}_{\bf{1}}}{\bf{ + 2}}{{\bf{a}}_{\bf{2}}}{\bf{x + }}...{\bf{ + n}}{{\bf{a}}_{\bf{n}}}{{\bf{x}}^{{\bf{n - 1}}}}{\bf{ + }}...{\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {\bf{2}} \right)\)
Substitute \({\bf{x = 0}}\)in the above expression,
\({{\bf{a}}_{\bf{1}}}{\bf{ = 0}}\)
From equation (2),
\({\bf{2}}{{\bf{a}}_{\bf{2}}}{\bf{x + }}...{\bf{ + n}}{{\bf{a}}_{\bf{n}}}{{\bf{x}}^{{\bf{n - 1}}}}{\bf{ + }}...{\bf{ = 0}}\)
Take the derivative with respect to x,
\({\bf{2}}{{\bf{a}}_{\bf{2}}}{\bf{ + }}...{\bf{ + n}}\left( {{\bf{n - 1}}} \right){{\bf{a}}_{\bf{n}}}{{\bf{x}}^{{\bf{n - 2}}}}{\bf{ + }}...{\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {\bf{3}} \right)\)
Substitute \({\bf{x = 0}}\)in the above expression,
\({{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\)
Therefore, from this process, we can show that:
\({{\bf{a}}_{\bf{i}}}{\bf{ = 0}}\)
Thus, this set forms the linearly independent set on every open interval (a, b).