1
$\begingroup$

I seem to be stuck defining an alternating sequence of terms in this series because $f^{(0)}(x)=f(x)$ is positive, as well as $f'(x)$, but then every other term starting with $f''(x)$ is negative. How can I define $f^{(n)}(x)$ given this?

\begin{array}{ll} f(x)=x^{\frac{1}{2}} & f(1)=1 \\ f'(x)=\frac{1}{2}\cdot x^{-\frac{1}{2}} & f'(1)=\frac{1}{2} \\ f''(x)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2}\cdot x^{-\frac{3}{2}} & f''(1)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2} \\ f'''(x)=(-1)^2\cdot 3\cdot\left(\frac{1}{2}\right)^{3}\cdot x^{-\frac{5}{2}} & f'''(1)=(-1)^2\cdot3\cdot\left(\frac{1}{2}\right)^{3} \\ f^{(4)}(x)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4}\cdot x^{-\frac{7}{2}} & f^{(4)}(1)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4} \\ f^{(n)}(x)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}\cdot x^{\frac{1-2n}{2}} & f^{(n)}(1)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n} \end{array}

I thought I had the right answer until I realized that I'd be defining $f(x)$ to be negative.

$\endgroup$
  • 2
    $\begingroup$ Please check your differentiation. $\endgroup$ – Tunococ Jul 13 '13 at 5:19
  • $\begingroup$ The power of $x$ changes as you take more derivatives so you should be getting things like $\frac 12 \frac 12 \frac 32 \frac 52 \dots$ with the appropriate sign. $\endgroup$ – Mark Bennet Jul 13 '13 at 5:20
  • $\begingroup$ @MarkBennet I don't see where I made an error? The powers of $x$ look proper to me. $\endgroup$ – Mirrana Jul 13 '13 at 5:23
  • $\begingroup$ Oh, ok.. I see an error with my coefficients, but it doesn't solve my original question... I'll fix that now. $\endgroup$ – Mirrana Jul 13 '13 at 5:25
  • $\begingroup$ @Tunococ I've fixed the differentiation, but now I'm stuck on how to define the sequence of odd numbers multiplied together. It's similar to $n!$ but it's not... $\endgroup$ – Mirrana Jul 13 '13 at 5:31
3
$\begingroup$

This is a formula which won't display so well in a comment.

$$1\cdot3\cdot5\cdot7=\frac {1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}{2\cdot4\cdot6}=\frac {7!}{2^33!}$$ You should be able to work out the general term from there.

Note also that there is no reason that every term of the sum has to fit the same neat formula. You can always write it as $$a_0+\sum_{r=1}^\infty a_rx^r$$ where $a_0$ is the term which does not fit the pattern, and $a_r$ has a general form.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why the power of 3 in the deominator? Wouldn't $2(3!)=2(1\cdot 2\cdot 3)=2\cdot 4\cdot 6$? $\endgroup$ – Mirrana Jul 13 '13 at 5:51
  • $\begingroup$ @agent154 - $2\cdot4\cdot6=(2\cdot1)\cdot(2\cdot2)\cdot(2\cdot3)$ - each term adds one to the factorial, and an extra factor of two. $2\times6=12$, $2\cdot4\cdot6=48=8\times6$. $\endgroup$ – Mark Bennet Jul 13 '13 at 7:59
0
$\begingroup$

Well you're not defining $f(x)$ to be negative, you're finding that certain derivatives of $f(x)$ are negative - if you think about the graph of this $f(x)$ it decreasingly increases; that is, it is increasing, but the rate at which it increases is decreasing, so it should seem to me that in the least, the second derivative should be negative, which is what your formula suggests!

Was in comments then realized this was a long comment :P

EDIT: So thanks to my oversight I wasn't too helpful. Hopefully this edit will provide more help

So if $f(x) = x^{1/2}$ then... $$ f'(x) = \frac{1}{2} x^{-1/2} \\ f''(x) = \frac{-1}{4} x^{-3/2} \\ f'''(x) = \frac{3}{8} x^{-5/2} \\ f^{(4)} (x) = \frac{-15}{16} x^{-7/2} \\ f^{(5)} (x) = \frac{105}{32} x^{-9/2} $$ so, at least to me, it appears as though you get $f^{(0)} (x) = x^{1/2}$, $f^{(1)} (x) = \frac{1}{2} x^{-1/2}$ and $f^{(n)} (x) = (-1)^{n+1} \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^n} x^{\frac{1-2n}{2}} : n \ge 2$. At least for me (I can be wrong though) I don't see any outstanding pattern other than the one I just said. Hope this helps!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But if I plug $n=0$ into the general formula, I get $f^{(0)}(x)=-\sqrt{x}$... and that doesn't match the given equation. $\endgroup$ – Mirrana Jul 13 '13 at 5:20
  • $\begingroup$ Oop! That would be my oversight, I apologize - I'll edit my answer $\endgroup$ – DanZimm Jul 13 '13 at 5:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.