Q28E

Question

Show that the set of functions\(\left\{ {{\bf{1,}}\,{\bf{cosx,}}\,{\bf{sinx,}}...{\bf{,}}\,{\bf{cosnx,}}\,{\bf{sinnx}}} \right\}\), 

where n is a positive integer is linearly independent on every open interval (a, b). Prove this in the special case n = 2 and (a, b) = \(\left( {{\bf{ - }}\infty {\bf{,}}\,\infty } \right)\)

 

Step-by-Step Solution

Verified
Answer

Therefore this set is from the linearly independent set on every open interval\(\left( {{\bf{ - }}\infty {\bf{,}}\,\infty } \right)\). Step-by-Step Solution

1Step 1: Use the fact that a polynomial of degree at most n has no more than n zeros unless it is identically zero.

A polynomial of degree n,

\({{\bf{a}}_{\bf{0}}}{\bf{1 + }}{{\bf{a}}_{\bf{1}}}\left( {{\bf{cosx + sinx}}} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{cos2x + sin2x}}} \right){\bf{ + }}...{\bf{ + }}{{\bf{a}}_{\bf{n}}}\left( {{\bf{cosnx + sinnx}}} \right){\bf{ = 0}}\)

We need to prove for n = 2, 

\({{\bf{a}}_{\bf{0}}}{\bf{1 + }}{{\bf{a}}_{\bf{1}}}\left( {{\bf{cosx + sinx}}} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{cos2x + sin2x}}} \right){\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{1}} \right)\)

Substitute \({\bf{x = 0}}\)in the above expression,

\(\begin{array}{c}{{\bf{a}}_{\bf{0}}}{\bf{1 + }}{{\bf{a}}_{\bf{1}}}\left( {{\bf{cos}}\left( {\bf{0}} \right){\bf{ + sin}}\left( {\bf{0}} \right)} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{cos2}}\left( {\bf{0}} \right){\bf{ + sin2}}\left( {\bf{0}} \right)} \right){\bf{ = 0}}\\{{\bf{a}}_{\bf{0}}}{\bf{ + }}{{\bf{a}}_{\bf{1}}}{\bf{ + }}{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{2}} \right)\end{array}\)

Take the derivative with respect to x, in equation (1),

\({{\bf{a}}_{\bf{1}}}\left( {{\bf{ - sinx + cosx}}} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{ - 2sin2x + 2cos2x}}} \right){\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{3}} \right)\)

Substitute \({\bf{x = 0}}\)in the above expression,

\(\begin{array}{c}{{\bf{a}}_{\bf{1}}}\left( {{\bf{ - sin}}\left( {\bf{0}} \right){\bf{ + cos}}\left( {\bf{0}} \right)} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{ - 2sin2}}\left( {\bf{0}} \right){\bf{ + 2cos2}}\left( {\bf{0}} \right)} \right){\bf{ = 0}}\\{{\bf{a}}_{\bf{1}}}{\bf{ + 2}}{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{4}} \right)\end{array}\)

Take the derivative with respect to x, in equation (3),

\({{\bf{a}}_{\bf{1}}}\left( {{\bf{ - cosx - sinx}}} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{ - 4cos2x - 4sin2x}}} \right){\bf{ = 0}}\)

Substitute \({\bf{x = 0}}\)in the above expression,

\(\begin{array}{c}{{\bf{a}}_{\bf{1}}}\left( {{\bf{ - cos}}\left( {\bf{0}} \right){\bf{ - sin}}\left( {\bf{0}} \right)} \right){\bf{ + }}{{\bf{a}}_{\bf{2}}}\left( {{\bf{ - 4cos2}}\left( {\bf{0}} \right){\bf{ - 4sin2}}\left( {\bf{0}} \right)} \right){\bf{ = 0}}\\{\bf{ - }}{{\bf{a}}_{\bf{1}}}{\bf{ - 4}}{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{5}} \right)\end{array}\)

2Step 2: Find the value of \({{\bf{a}}_{\bf{0}}}{\bf{,}}\,{{\bf{a}}_{\bf{1}}}\) and \({{\bf{a}}_{\bf{2}}}\),

Solve the equation (4) and (5),

 

\(\begin{array}{c}{{\bf{a}}_{\bf{1}}}{\bf{ + 2}}{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\\{\bf{ - }}{{\bf{a}}_{\bf{1}}}{\bf{ - 4}}{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\\{\bf{ - 2}}{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\\{{\bf{a}}_{\bf{2}}}{\bf{ = 0}}\end{array}\)

 

Substitute the value of \({{\bf{a}}_{\bf{2}}}\)  in the equation (4),

\(\begin{array}{c}{{\bf{a}}_{\bf{1}}}{\bf{ + 2}}\left( {\bf{0}} \right){\bf{ = 0}}\\{{\bf{a}}_{\bf{1}}}{\bf{ = 0}}\end{array}\)

 

Substitute the value of \({{\bf{a}}_{\bf{1}}}\) and \({{\bf{a}}_{\bf{2}}}\)  in the equation (2),

 

\(\begin{array}{c}{{\bf{a}}_{\bf{0}}}{\bf{ + }}\left( {\bf{0}} \right){\bf{ + }}\left( {\bf{0}} \right){\bf{ = 0}}\\{{\bf{a}}_{\bf{0}}}{\bf{ = 0}}\end{array}\)

3Step 3: conclusion, Step 3: conclusion,

Therefore, from this process, we can show that 

 

\({{\bf{a}}_{\bf{0}}}{\bf{ = 0,}}\,{{\bf{a}}_{\bf{1}}}{\bf{ = 0,}}\,{{\bf{a}}_{\bf{2}}}{\bf{ = 0,}}\,\)

 

Therefore this set is from the linearly independent set on every open interval\(\left( {{\bf{ - }}\infty {\bf{,}}\,\infty } \right)\).