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There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

Added see Is it true that all the "irrational power" functions are almost polynomial ?Is it true that all the "irrational power" functions are almost polynomial ? .

There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

Added see Is it true that all the "irrational power" functions are almost polynomial ? .

There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

Added see Is it true that all the "irrational power" functions are almost polynomial ? .

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Robin Chapman
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There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

Added see Is it true that all the "irrational power" functions are almost polynomial ? .

There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.

Added see Is it true that all the "irrational power" functions are almost polynomial ? .

Source Link
Robin Chapman
  • 20.8k
  • 2
  • 66
  • 81

There's a generalization of the mean value theorem that can be applied here, namely that if $f$ is smooth enough then for each $x$ there is $y$ such that $$\Delta^kf(x)=f^{(k)}(y)$$ and $x < y < x+k$.