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Define a function $f: \Bbb{R} \to \Bbb{R}$ by $f(x)=\displaystyle \sum\limits_{n=0}^{\infty}(-1)^n\frac{x^n}{n!n!}$

Show that f is decreasing on on $x \in [0,2]$ and that there exists a unique $x_0 \in [0,2]$ for which $f(x_0)=0$.

I started by calculating $f'$ but I do not know what to do next.

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    $\begingroup$ Hints: (a) if $f$ is decreasing, what does that imply for $f'$ (and vice versa)? (b) Can you calculate $f(0)$? Can you calculate $f(2)$? (You may not want to calculate $f(2)$ explicitly, but a clever argument may be able to bound its value; if you're familiar with the bounds on decreasing alternating series, that will help here.) $\endgroup$ Jan 14, 2013 at 17:27
  • $\begingroup$ @stevenStadnicki I have tried proving $f'<0$ but I have no clue about what I should do. As for (b) If $f(0) = 0$ and the function is decreasing wouldn't that mean that $x_0$ doesn't exist? $\endgroup$
    – user10444
    Jan 14, 2013 at 17:34
  • $\begingroup$ If $f(0)$ were $0$ and $f$ was decreasing, that would give you $x_0=0$ - but you may wish to recalculate $f(0)$; in particular, what's the $n=0$ term? $\endgroup$ Jan 14, 2013 at 17:41
  • $\begingroup$ @stevenStadnicki I assumed $0^0=0$ so if $f(1)=1$ and $f(2)$ can be approximated and is negative then $x_0$ exists. Thank you $\endgroup$
    – user10444
    Jan 14, 2013 at 17:47
  • $\begingroup$ $0^0$ is canonically taken to be $1$ for power series expansions - this means that the constant term is $(-1)^0\frac{x^0}{0!0!} = 1$ regardless of what $x$ is (it's a poor constant term that changes based on the value of a variable!). $\endgroup$ Jan 14, 2013 at 18:17

1 Answer 1

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The derivative is of the form $\sum (-1)^n a_n(x)$, where you can prove that $0 \leq a_{n+1}(x) \leq a_n(x)$ for every $n \geq 1$ and $x \in [0,2]$. Such an alternating series has the same sign as its first term (prove it!).


In order to prove that $f(2) < 0$, you can start noticing that $f(2) = 1 - 2 + 1 + R$ with $$ R = \sum_{n=3}^\infty (-1)^n\frac{2^n}{(n!)^2} $$ and justify why $R < 0$.

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  • $\begingroup$ Can it be done by using approximation? $\endgroup$
    – user10444
    Jan 14, 2013 at 17:53
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    $\begingroup$ I think it is better not to use approximation. $\endgroup$
    – Siméon
    Jan 14, 2013 at 18:05
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    $\begingroup$ @user10444, write out the first few terms of the derivative as a power series. Then pair up the first term with the second in parentheses, the third with the fourth, the fifth with the sixth, and so on. Each pair has the same resulting $\pm$ sign when $x$ is restricted, in your case to $0 \leq x \leq 2.$ $\endgroup$
    – Will Jagy
    Jan 14, 2013 at 18:09

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