Boundedness of $n^{th}$ derivative

Question

Let $f(x) =\arctan(x) $ then the sequence {$f^n(0) $} is bounded for all positive even integers n ?

What i have think denominator of first derivative you can see that i will never attain zero moreover which is a cts function and any nth derivative will be of higher order than the numerator .. so it will be bounded

What is a cts function? If this means continuous, please take the time to write out the word. Otherwise, you should explain what it means. — Théophile, Jun 21 '17 at 17:38

farruhota · Accepted Answer · 2017-06-21T19:08:40.473

0

Expand to Taylor series: $$y=\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$ $$y'=1-x^2+x^4-x^6+\cdots$$ $$y''=-2x+4x^3-6x^5+\cdot$$ $$y'''=-2+12x^2-30x^4+\cdots$$ $$\cdots$$

Thus: $$y^{(2n)}(0)=0, n\ge 2$$ $$y^{(2n+1)}(0)=(-1)^n(2n)!, n\ge 1$$

edited Jun 21 '17 at 19:08

answered Jun 21 '17 at 18:13

farruhota

32,168

1

You should have a $(-1)^n$ factor in the last line. – Bernard Jun 21 '17 at 18:22
@Bernard, thank you. Corrected. – farruhota Jun 21 '17 at 19:09

score 0 · Answer 2 · answered Jun 21 '17 at 18:28

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HINT: Derivative of odd function is even and derivative of even function is odd,for the proof of this take a look at this

answered Jun 21 '17 at 18:28

kingW3

13,734

Boundedness of $n^{th}$ derivative

2 Answers2