I would like to have some verification to see if my answer is correct. The given function is $f(x)=ln(1+x^2)$ and I need the 40th derivative at $x=0$. Here is my work: Using series one can manipulate $\frac{1}{1+x}=1-x+x^2-x^3+x^4...=SUMx^n(-1)^n$ into $\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8...=SUMx^{2n}(-1)^n$. Then $\frac{2x}{1+x}=2x-2x^3+2x^5-2x^7+2x^9...=2SUMx^{2n+1}(-1)^n$. Integrating gives $ln(1+x^2)=...=2SUM\frac{x^{2n+2}(-1)^n}{2n+2}$ where a $2$ cancels to arrive at a $n+1$ in that denominator. Now for the 40th derivative, $2n+2=40$ gives $n=19$ and thus I believe the answer is $\frac{-40!}{20}$ Do you concur? If not could you correct me? Thanks.
$\begingroup$
$\endgroup$
17
-
2$\begingroup$ Why a downvote without an explanation? $\endgroup$– imranfatApr 30, 2015 at 20:53
-
2$\begingroup$ @Zach466920 I am posting my work with my question. If people post questions without any effort, then naturally the bloggers are not going to do the work for the OP. But I am showing my work and I would like to be sure that what I am doing is right. What's unsuitable about that? The worst you can say is about my poor formatting for which I apologize $\endgroup$– imranfatApr 30, 2015 at 20:58
-
1$\begingroup$ @Zach466920 your comment is not relevent imo. downvoting this question is silly. $\endgroup$– mookidApr 30, 2015 at 21:11
-
3$\begingroup$ @Zach466920: Wait--what? Are you saying that if the question can be answered in one sentence, it should be closed instead of answered? $\endgroup$– MPWApr 30, 2015 at 21:38
-
1$\begingroup$ @Zach466920: In view of some of the gibberish that slops onto this forum, I would say this is a welcome relief and a fine example of a well-posed question. It exhibits evidence of effort, a clear explanation of what was attempted, a markedly mature grasp of style, and a complete absence of the meandering guesswork that one often sees. If only we saw this more often! $\endgroup$– MPWApr 30, 2015 at 23:34
|
Show 12 more comments
1 Answer
$\begingroup$
$\endgroup$
3
Taking the $40$th derivative at 0 will yield you the coefficient for the $x^{40}$ term multiplied by $40!$
The general series for natural log is:
$$ ln(x+1) = \frac {x^1}1 - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4... $$
replacing $x^2$ makes:
$$ ln(x^2+1) = \frac {x^2}1 - \frac {x^4}2 + \frac {x^6}3 - \frac {x^8}4... $$
From this it is apparent that the 20th term of the first series shares its coefficient with the 40th term of the second series.
The coefficient equals $-\frac 1{20}$. Multiplying $40!$ to cancel the implied division yields: $-\frac {40!}{20}$
From this I confirm your answer to be correct; however, your proof and solution is incredibly messy. There is a known series for natural log. Using a fractional series and integrating was a waste of time and effort.
answered Nov 28, 2015 at 19:10
-
$\begingroup$ Hmmm, first taking an anti derivative and then replacing the $x$ by $x^2$ is indeed much faster. Thanks for the reply. $\endgroup$– imranfatNov 28, 2015 at 19:30
-
$\begingroup$ Yes, but that ln series came from the geometric series... $\endgroup$– imranfatNov 28, 2015 at 19:59
-
$\begingroup$ Indeed it was discovered that way. Actually quite a lot of series stuff (and sums) was discovered after "exhausting" possibilities of what one can do with geometric series. I remember I learned the series of $ln(1+x)$ as well as $ln(1-x)$ back in the 17th/18th century that way. I do agree that all these standard series ought to be remembered by heart, but that counts for series for $sinx$ , $cosx$ and all the other "standard" transcendental functions as well. I guess I am a dinosaur here, communicating with a great duck :) $\endgroup$– imranfatNov 28, 2015 at 23:08