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badreferences Group Title

Let \(n\in\mathbb N\). For\[e^xf_n(x)=\sum_{k=1}^\infty\frac{k^nx^k}{\left(k-1\right)!}\]show that \(f_n(x)\) is a polynomial of degree \(n+1\) with integer coefficients. Tricky question.

  • 2 years ago
  • 2 years ago

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  1. mukushla Group Title
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    take derivative\[e^x=\sum_{k=1}^\infty\frac{x^{k-1}}{(k-1)!}\]multiply by x\[xe^x=\sum_{k=1}^\infty\frac{x^{k}}{(k-1)!}\]and take derivative like this n times and multiply by x again every time finally we will have something like this\[e^xP(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}\]

    • 2 years ago
  2. mukushla Group Title
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    actually better to write\[e^xP_n(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}\]

    • 2 years ago
  3. mukushla Group Title
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    lets prove by induction that degree of \(P_n(x)\) is n+1 \[P_1(x)=x+x^2\]lets say degree of \(P_n(x)\) is n+1 and prove that degree of \(P_{n+1}(x)\) is n+2\[P_{n+1}(x)=xe^{-x}[e^xP_n(x)]'=x(P_n^'(x)+P_n(x))\] which is from degree 1+n+1=n+2

    • 2 years ago
  4. sauravshakya Group Title
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    @mukushla can u show me the derivative of |dw:1348828878035:dw|

    • 2 years ago
  5. mukushla Group Title
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    \[e^x(1+x)=\sum_{k=1}^\infty\frac{kx^{k-1}}{(k-1)!}\]

    • 2 years ago
  6. sauravshakya Group Title
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    OH....... NOW, I got it.... THANX @mukushla

    • 2 years ago
  7. badreferences Group Title
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    http://openstudy.com/study#/updates/50650f91e4b08d185211d261 Solution in last post.

    • 2 years ago
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