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Forget about the $\cos,\sin$ function, show that $\left|1-x^2/2!+x^4/4!-x^6/6!+...\right|\leq1$

I tried to use differentiation, but it doesn't seems helpful. Please help. Thanks.

JSCB
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Define $$f(x)=\sum_{n\geqslant 0} (-1)^n\frac{x^{2n}}{(2n)!}$$

One can readily see this converges for any $x$. Note that $$f''+f=0$$ whence $$f'f''+ff'=0$$ which means $$f'^2+f^2=K$$

By plugging in values we see $K=1$; thus $$f'^2+f^2=1$$ whence we must have $|f'|,|f|\leq 1$ for all $x$.

Pedro
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  • Readily because it's alternating and eventually strictly decreasing in absolute value? – dfeuer Sep 01 '13 at 02:54
  • @dfeuer It is absolutely convergent. – Pedro Sep 01 '13 at 02:55
  • You could use the ratio test to prove absolute convergence or just quickly use the AST for conditional convergence. – Jon Claus Sep 01 '13 at 02:55
  • I just realized the ratio test does it. – dfeuer Sep 01 '13 at 02:57
  • The weird thing, to me, is that the first several terms can actually get large for $|x|\gg1$, but then things have to cancel out somehow. – dfeuer Sep 01 '13 at 03:00
  • @dfeuer The point is that for any $x$ we can use up enough terms to get the value inside $[-1,1]$- – Pedro Sep 01 '13 at 03:01
  • Yes, you've proven that, and I know it works; it's just "feels" weird. – dfeuer Sep 01 '13 at 03:03
  • A small remark. From the first equation in Peter's solution, $|f''| \leq 1$ for all $x$ also. By putting $y(x)=f'(x)$ and following the same steps leads to $|f^{(k)}| \leq 1$ for all $x$. – Max Sep 01 '13 at 03:10