Collection of Maths Problems

Generating functions for given sequences

Task number: 3868

• Variant

  \((1{,}1,1{,}1,1{,}1,\ldots)\)

• Solution

  The sequence \((1{,}1,1{,}1,1{,}1,\ldots)\) corersponds to the series \(1+x+x^2+x^3+…\) which has around 0 sum \(\frac{1}{1-x}\).

• Answer

  The generating function for the sequence \((1{,}1,1{,}1,1{,}1,\ldots)\) is \(\frac{1}{1-x}\).

• Variant

  \((1,-1{,}1,-1{,}1,-1,\ldots)\)

• Solution

  We use the generating function for \((1{,}1,1,…)\) and substitute \(-x\) for \(x\).

• Answer

  The generating function for the sequence \((1,-1{,}1,-1{,}1,-1,\ldots)\) is \(\frac{1}{1+x}\).

• Variant

  \((1{,}0,1{,}0,1{,}0,\ldots)\)

• Solution

  We use the generating function for \((1{,}1,1,…)\) and substitute \(x^2\) for \(x\).

• Answer

  The generating function for the sequence \((1{,}0,1{,}0,1{,}0,\ldots)\) is \(\frac{1}{1-x^2}\).

• Variant

  \((1{,}2,3{,}4,5{,}6,\ldots)\)

• Solution

  We use the generating function for \((1{,}1,1,…)\) and derive it.

• Answer

  The generating function for the sequence \((1{,}2,3{,}4,5{,}6,\ldots)\) is \(\frac{1}{(1-x)^2}\).

• Variant

  \((1,-1{,}2,-2{,}3,-3{,}4,-4,\ldots)\)

• Solution

  We split the sequence \((1,-1{,}2,-2{,}3,-3{,}4,-4,\ldots)\) into the sum of \((1{,}0,2{,}0,3{,}0,4{,}0,\ldots)\) and \((0,-1{,}0,-2{,}0,-3{,}0,-4,\ldots)\).

  We obtain the generating function for the first sequence by the substitution \(x^2\) for \(x\) in the generating function for the sequence \((1{,}2,3{,}4,\ldots)\).

  The other is obtained by a further product with \(-x\).

• Answer

  The generating function for the sequence \((1,-1{,}2,-2{,}3,-3{,}4,-4,\ldots)\) is \(\frac{1-x}{(1-x^2)^2}\).

• Variant

  \((2{,}1,4{,}3,6{,}5,\ldots)\)

• Solution

  We split the sequence \((2{,}1,4{,}3,6{,}5,\ldots)\) into the sum of \((1{,}2,3{,}4,5{,}6,\ldots)\) and \((1,-1{,}1,-1{,}1,-1,\ldots)\). Their generating functions are already known.

• Answer

  The generating function for the sequence \((2{,}1,4{,}3,6{,}5,\ldots)\) is \(\frac{1}{(1-x)^2}+\frac{1}{1+x}\).

• Variant

  \((1{,}2,1{,}4,1{,}8,\ldots)\)

• Solution

  We split the sequence \((1{,}2,1{,}4,1{,}8,\ldots)\) into the sum of \((1{,}0,1{,}0,1{,}0,\ldots)\) and \((0{,}2,0{,}4,0{,}8,\ldots)\).

  The generating function for the firs is already known. For the other we start with \((1{,}2,4{,}8,\ldots)\), substitute \(x^2\) and multiply by \(2x\).

• Answer

  The generating function for the sequence \((1{,}2,1{,}4,1{,}8,\ldots)\) is \(\frac{1}{1-x^2}+\frac{2x}{1-2x^2}\).

• Variant

  \((1,-3{,}5,-7{,}9,-11,\ldots)\)

• Solution

  The sequnece \((1{,}2,3{,}4,5{,}6,\ldots)\) multiply by two, subtract \((1{,}1,1{,}1,1{,}1,\ldots)\) and get \((1{,}3,5{,}7,9{,}11,\ldots)\). Now it suffices to substitute \(-x\) for \(x\).

• Answer

  The generating function for the sequence \((1,-3{,}5,-7{,}9,-11,\ldots)\) is \(\frac{2}{(1+x)^2}-\frac{1}{1+x}=\frac{1-x}{(1+x)^2}\).

• Variant

  \((0{,}1,1{,}2,2{,}4,3{,}8,4{,}16,\ldots,i,2^i,\ldots)\)

• Solution

  We split the sequence \((0{,}1,1{,}2,2{,}4,3{,}8,4{,}16,\ldots,i,2^i,\ldots)\) into the sum of \((0{,}0,1{,}0,2{,}0,3{,}0,4{,}0,\ldots,i,0,\ldots)\) and \((0{,}1,0{,}2,0{,}4,0{,}8,0{,}16,\ldots,2^i,0,\ldots)\).

  The generating function for the first one we get, if we subtitute into the generating function for \((1{,}2,3{,}4,\ldots)\) term \(x^2\) for \(x\), and then multiply by \(x^2\).

  The other we get if we use the generating function for \((1{,}1,1,\ldots)\), substitute \(2x\) for \(x\), then substitute \(x^2\) for \(x\), and finally multiply by \(x\).

• Answer

  The generating function for the sequence \((0{,}1,1{,}2,2{,}4,3{,}8,4{,}16,\ldots,i,2^i,\ldots)\) is \(\frac{x^2}{(1-x^2)^2}+\frac{x}{1-2x^2}\).