User:Egm4313.s12.team8.lucas/R5.2

=R5.2=

Problem Statement
Determine whether the following pairs of functions are linearly independent using the Wronskian. Then use the Gramian to verify the result over the interval $$\ [a,b]=[-1,1]$$

1) $$\ f(x)=x^2, g(x)=x^4$$

2) $$\ f(x)=cosx, g(x)=sin3x$$

Part 1
$$\ f(x)=x^2, g(x)=x^4$$

The Wronskian of two functions, f and g, is defined as:

$$\ W(f,g)=\det \begin{bmatrix} f & g \\f' & g' \end{bmatrix}$$

$$\ f(x)=x^2 \Rightarrow f'(x)=2x$$

$$\ g(x)=x^4 \Rightarrow g'(x)=4x^3$$

By substituting functions the Wronskian becomes:

$$\ W(x^2,x^4)=\det \begin{bmatrix} x^2 & x^4 \\2x & 4x^3 \end{bmatrix}$$

$$\ W(x^2,x^4)=(x^2)(4x^3)-(x^4)(2x)=4x^5-2x^5=2x^5 \ne 0$$

The Wronskian does not equal 0, therefore the two functions are linearly independent.

The Gramian is defined as:

$$\ \Gamma(f,g)=\det \begin{bmatrix} \langle f,f \rangle & \langle f,g \rangle \\ \langle g,f \rangle & \langle g,g \rangle \end{bmatrix}$$

where:

$$\ \langle f,g \rangle=\int_a^b f(x)g(x)\ dx$$

For the above functions:

$$\ \langle x^2,x^2 \rangle=\int_{-1}^1 (x^2)(x^2)\ dx=\int_{-1}^1 x^4\ dx=\frac{x^5}{5} \mid_{-1}^1=\frac{2}{5}$$

$$\ \langle x^2,x^4 \rangle=\int_{-1}^1 (x^2)(x^4)\ dx=\int_{-1}^1 x^6\ dx=\frac{x^7}{7} \mid_{-1}^1=\frac{2}{7}$$

$$\ \langle x^4,x^2 \rangle=\int_{-1}^1 (x^4)(x^2)\ dx=\int_{-1}^1 x^6\ dx=\frac{x^7}{7} \mid_{-1}^1=\frac{2}{7}$$

$$\ \langle x^4,x^4 \rangle=\int_{-1}^1 (x^4)(x^4)\ dx=\int_{-1}^1 x^8\ dx=\frac{x^9}{9} \mid_{-1}^1=\frac{2}{9}$$

The Gramian becomes:

$$\ \Gamma(x^2,x^4)=\det \begin{bmatrix} 2/5 & 2/7 \\ 2/7 & 2/9 \end{bmatrix}$$

$$\ \Gamma=(\frac{2}{5})(\frac{2}{9})-(\frac{2}{7})(\frac{2}{7})=\frac{16}{2205} \ne 0$$

The Gramian does not equal 0, therefore the functions are linearly independent.

Part 2
$$\ f(x)=cosx, g(x)=sin3x$$

The Wronskian of two functions, f and g, is defined as:

$$\ W(f,g)=\det \begin{bmatrix} f & g \\f' & g' \end{bmatrix}$$

$$\ f(x)=cosx \Rightarrow f'(x)=-sinx$$

$$\ g(x)=sin3x \Rightarrow g'(x)=3cos3x$$

By substituting functions the Wronskian becomes:

$$\ W(cosx,sin3x)=\det \begin{bmatrix} cosx & sin3x \\-sinx & 3cos3x \end{bmatrix}$$

$$\ W(cosx,sin3x)=(cosx)(3cos3x)-(sin3x)(-sinx)=3cosxcos3x+sinxsin3x \ne 0$$

The Wronskian does not equal 0, therefore the two functions are linearly independent.

The Gramian is defined as:

$$\ \Gamma(f,g)=\det \begin{bmatrix} \langle f,f \rangle & \langle f,g \rangle \\ \langle g,f \rangle & \langle g,g \rangle \end{bmatrix}$$

where:

$$\ \langle f,g \rangle=\int_a^b f(x)g(x)\ dx$$

For the above functions:

$$\ \langle cosx,cosx \rangle=\int_{-1}^1 (cosx)(cosx)\ dx=\int_{-1}^1 cos^{2}x\ dx=\int_{-1}^1 \frac{1+cos2x}{2}\ dx=x/2+sin(2x)/4 \mid_{-1}^1=1.455$$

$$\ \langle cosx,sin3x \rangle=\int_{-1}^1 (cosx)(sin3x)\ dx=\frac{1}{2}\int_{-1}^1 [sin4x-sin(-2x)]\ dx=\frac{1}{2}[\frac{-cos4x}{4}-\frac{cos(-2x)}{2}] \mid_{-1}^1=0$$

$$\ \langle sin3x,cosx \rangle=\int_{-1}^1 (sin3x)(cox)\ dx=\frac{1}{2}\int_{-1}^1 [sin4x-sin(-2x)]\ dx=\frac{1}{2}[\frac{-cos4x}{4}-\frac{cos(-2x)}{2}] \mid_{-1}^1=0$$

$$\ \langle sin3x,sin3x \rangle=\int_{-1}^1 (sin3x)(sin3x)\ dx=\int_{-1}^1 sin^{2}3x\ dx=\int{-1}^1 \frac{1-cos6x}{2}\ dx=x/2-sin(6x)/12 \mid_{-1}^1=1.0465$$

The Gramian becomes:

$$\ \Gamma(cosx,sin3x)=\det \begin{bmatrix} 1.455 & 0 \\ 0 & 1.0465 \end{bmatrix}$$

$$\ \Gamma=(1.455)(1.0465)-(0)(0)=1.522 \ne 0$$

The Gramian does not equal 0, therefore the functions are linearly independent.