BIM_XY's blog

Help me!
By BIM_XY, history, 7 months ago, In English

We know $$$f(x)\equiv \exp(\ln f(x))\equiv \pmod Q$$$ so $$$f^k(x) \equiv \exp(k\ln f(x))\equiv \exp(k\bmod Q\ln f(x))\equiv f^{k\bmod Q} (x) \pmod Q$$$ (Here, it is assumed that $$$\exp$$$ and $$$\ln$$$ are polynomials), but when $$$f(x)=(x+1),Q=2,k=2$$$, $$$f^k(x)\equiv 1\pmod Q$$$, but $$$f^{k\bmod Q}(x)\equiv x^2+1\pmod Q$$$, WHY??

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7 months ago, hide # |
 
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Because i'm weak in maths. ;)

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7 months ago, hide # |
 
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$$$\exp$$$ and $$$\ln$$$ aren't well defined as polynomials in $$$\mathbb Z/Q\mathbb Z$$$ because their taylor series contain coefficients $$$\frac1{Q!}$$$ and $$$\frac1Q$$$ respectively, which mean division by 0 when working mod Q.

Also even without the mod Q, $$$\ln x$$$ does not have a power series representation centered at 0, the radius of convergence is 0.

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