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Let $(f_n)$ be a sequence of functions with $$f_n(x)=\sum_{k=0}^n (-1)^k \frac{x^{2k}}{(2k)!}.$$ Find a function $f:[0,1] \to \Bbb R$ such that $f_n \to f$ on $[0,1]$.

I know that this sequence of function is the partial sum of the Maclaurin series of $\cos x$, and so, $f_n \to f$ on $[0,1]$, where $f(x)=\cos x$. But, the codomain of $f$ is not $\Bbb R$, but $[-1,1]$. How to approach this?

Thanks in advanced.

lap lapan
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    Does $f: [0, 1] \to \mathbb{R}$ means that the codomain of $f$ is $\mathbb{R}$? I think it only means that the codomain is contained in $\mathbb{R}$, so you can just take $f(x) = \cos x$. – Seewoo Lee May 15 '22 at 03:18
  • The codomain is not the same as the image. See here. So long as for each $x \in X$, we have $f(x) \in Y$, then $f$ is a function $X \to Y$, which is true in this case. I think Seewoo Lee's comment is confusing because $f: [0, 1] \to \Bbb R$ absolutely does mean that the codomain is $\Bbb R$, it just doesn't mean that $\Bbb R$ is the image. – Izaak van Dongen May 17 '22 at 14:27

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