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Let $f:\mathbb{R} \rightarrow \mathbb{R} $ be a non-constant, three times differentiable function. If $f(1+\frac{1}{n})=1$ for all integers $n$, then find $f''(1)$.

I could find $f'(1)=\lim_{n \rightarrow \infty} \frac{f(1+\frac{1}{n})-f(1)}{\frac{1}{n}}$. Now, $f(1)=1$ due to continuity of $f$ at $1$. But the expression of $f''(1)$ is becoming overtly complicated. Please help.

user26857
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    Use $f(1+x)=f(1)+xf'(1)+\frac12x^2f''(1)+o(x^2)$ when $x\to0$ to identify $f(1)=1$ and $f'(1)=f''(1)=0$. Only twice differentiability (not thrice) at $1$ (not everywhere) is needed. – Did Feb 02 '19 at 15:43
  • @Did I think you need thrice diff. at $;x=1;$ to assure $;o(x^2);$ when $;x\to 0;$ . For the residue of Taylor's (or Maclaurin's) polynomial of degree two to converge to zero that way we need the third derivative. – DonAntonio Feb 02 '19 at 15:53
  • @DonAntonio No, you might be confusing the conditions for Taylor theorem to hold and the conditions to get some explicit formulas for its remainder term. It happens that WP is relatively precise on the topic, hence please consult this page. – Did Feb 02 '19 at 16:31

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Observe that since we know that $\;f''(1)\;$ exists, for $\;n\in\Bbb N\;$ , we have that (use for example Check my workings: Show that $\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)$)

$$f''(1)=\lim_{n\to\infty}\frac{f\left(1+\frac1n\right)-2f(1)+f\left(1-\frac1n\right)}{\left(\frac1n\right)^2}=\lim_{n\to\infty}\frac{1-2+1}{\frac1{n^2}}=0$$

Because, as you say, by continuity $\;f(1)=1\;$ .

Warning: One equality sign before the last one I am not passing to the limit in the numerator while keeping the denominator as it is. This would require justification. What I'm doing is substituting the actual values of $\;f\;$ at those points!

DonAntonio
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  • Nifty although this requires to know the behaviour of $f$ on the left of $1$, which is more than needed to conclude. The OP is ambiguous on this point but the exercice is usually given by assuming $f(1+\frac1n)=1$ for every natural integer $n$. – Did Feb 02 '19 at 16:33
  • @Did Perhaps, yet the OP wrote simply "integer", so left or right is just the same. – DonAntonio Feb 02 '19 at 16:45
  • Easy... nobody is attacking you, I just made two factual remarks. – Did Feb 02 '19 at 16:46
  • @ Did Ok...I didn't think you were attacking me, honest. I was just remarking what the question says. Thanks. – DonAntonio Feb 02 '19 at 16:58
  • Cool. $ $ $ $ $ $ – Did Feb 02 '19 at 16:58