Collection of Maths Problems

Independence of functions

Task number: 2521

• Variant 1

  \(\{2x-1,x-2, 3x\}\).

• Resolution

  Lets denote functions \(f(x)=2x-1, g(x)=x-2\) and \(h(x)=3x\).

  We seek for the coefficients \(a,b\) and \(c\), such that \(a\cdot f + b\cdot g + c \cdot h = 0\) holds, where on the right hand side, there is a zero-vector of this vector space, that is a constant function, that maps everything to a zero (in \(\mathbb R\)).

  Two functions in this space are equal if they are equal in all points, that is for every \(x\in \mathbb R\) it must hold that \(a(2x-1)+b(x-2)+c(3x)=(2a+b+3c)x+(-a-2b)=0\). This again lead to the homogenous system of equations \(2a+b+3c=0, -a-2b=0\).

  This system has a non-trivial solution, for example \((a,b,c)^T=(-2{,}1,1)^T\), and thus the set is linear dependent.

• Result

  \(\{2x-1,x-2, 3x\}\) is linear dependent set.

• Variant 2

  \(\{x^2+2x+3, x+1, x-1\}\).

• Result

  \(\{x^2+2x+3, x+1, x-1\}\) is linear independent set.

• Variant 3

  \(\{\sin(x), \cos(x)\}\).

• Resolution

  The equation \(a\sin(x)+b\cos(x)=0\) for \(x=0\) implies \(b=0\), similarly for \(x=\frac\pi2\) we obtain \(a=0\).

• Result

  \(\{\sin(x), \cos(x)\}\) is linear independent.

• Variant 4

  \(\{\sin(x+1), \sin(x+2)\, \sin(x+3)\}\).

• Hint

  Use the formula for \(\sin(x+y)\).

• Resolution

  The following holds
  \(\sin(x+1)=\sin(x)\cos(1)+\cos(x)\sin(1)\),
  \(\sin(x+2)=\sin(x)\cos(2)+\cos(x)\sin(2)\),
  \(\sin(x+3)=\sin(x)\cos(3)+\cos(x)\sin(3)\).

  Using these formulae we obtain the following \(a_1\sin(x+1)+a_2\sin(x+2)+a_3\sin(x+3)=\\ \Big(a_1\cos(1)+a_2\cos(2)+a_3\cos(3)\Big)\sin(x)+ \Big(a_1\sin(1)+a_2\sin(2)+a_3\sin(3)\Big)\cos(x)\).

  Now because \(\sin(x)\) and \(\cos(x)\) are linear independent (by the solution to the previous variant) the expression is equivalent to the zero-function if and only if

  \(a_1\cos(1)+a_2\cos(2)+a_3\cos(3)=0\) and
  \(a_1\sin(1)+a_2\sin(2)+a_3\sin(3)=0\).

  But this is a homogeneous system of equations, which consists of two equations with three variables, and thus there must be a non-trivial solution to the system.

  One particular solution is \((a_1,a_2,a_3)^T=\\\left( \cos(2)\sin(3)-\cos(3)\sin(2), \cos(3)\sin(1)-\cos(1)\sin(3), \cos(1)\sin(2)-\cos(2)\sin(1) \right)^T=\\\left(\sin(1),-\sin(2),\sin(1)\right)^T \)

• Result

  \(\{\sin(x+1), \sin(x+2)\, \sin(x+3)\}\) is linear dependent.

• Variant 5

  \(\{\ln(x),\log(2x),\log_2(x^2)\}\).

  (i.e. the natural, decadic and binary logarithm.)

• Resolution

  We can use the following formulae \(\log(2x)=\frac{\ln x+\ln 2}{\ln 10}\), \(\log_2(x^2)=\frac{2\ln x}{\ln 2}\).

  Now in the equation \(a_1\ln(x)+a_2 \log(2x)+ a_3 \log_2(x^2)=0\), we see that the first and the third terms are multiples of the another.

  The equation has a non-trivial solution, for example \((a_1,a_2,a_3)^T=(-2{,}0,\ln 2)^T\)