let $f(x) = tan ^{-1}x, x in mathbb{R} $ .then choose the correct statement












0












$begingroup$


let $f(x) = tan ^{-1}x, x in mathbb{R} $ .then



choose the correct statement



$1.$ there exist a polynomial $p(x)$ satisfying $p(x)f'(x) =1$ for all $x$



$2.$$f^{(n)}(0) =0$ for all positive even integer $n$



$3.$ the sequence ${f^{(n)} (0) }$ is unbounded



$4.$$f^{(n)} (0) =0$ for all $n$



My attempt : if $ x =0$ ,then obviously $f(x) =0$ because $tan 0 =0$



from this i see option $2)$ and option $4)$ will correct



other option i don't know...



Any hints/solution will be apprecaited



thanks u






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    For $(2)$ and $(4)$, I think you misunderstood $f^{(n)}(0)$ as $(f(0))^n; f^{(n)}(0)$ means the $n^{th}$ derivative of $f$ at $x=0$.
    $endgroup$
    – Shubham Johri
    Dec 12 '18 at 20:45












  • $begingroup$
    thanks @ShubhamJohri...
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:53
















0












$begingroup$


let $f(x) = tan ^{-1}x, x in mathbb{R} $ .then



choose the correct statement



$1.$ there exist a polynomial $p(x)$ satisfying $p(x)f'(x) =1$ for all $x$



$2.$$f^{(n)}(0) =0$ for all positive even integer $n$



$3.$ the sequence ${f^{(n)} (0) }$ is unbounded



$4.$$f^{(n)} (0) =0$ for all $n$



My attempt : if $ x =0$ ,then obviously $f(x) =0$ because $tan 0 =0$



from this i see option $2)$ and option $4)$ will correct



other option i don't know...



Any hints/solution will be apprecaited



thanks u






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    For $(2)$ and $(4)$, I think you misunderstood $f^{(n)}(0)$ as $(f(0))^n; f^{(n)}(0)$ means the $n^{th}$ derivative of $f$ at $x=0$.
    $endgroup$
    – Shubham Johri
    Dec 12 '18 at 20:45












  • $begingroup$
    thanks @ShubhamJohri...
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:53














0












0








0





$begingroup$


let $f(x) = tan ^{-1}x, x in mathbb{R} $ .then



choose the correct statement



$1.$ there exist a polynomial $p(x)$ satisfying $p(x)f'(x) =1$ for all $x$



$2.$$f^{(n)}(0) =0$ for all positive even integer $n$



$3.$ the sequence ${f^{(n)} (0) }$ is unbounded



$4.$$f^{(n)} (0) =0$ for all $n$



My attempt : if $ x =0$ ,then obviously $f(x) =0$ because $tan 0 =0$



from this i see option $2)$ and option $4)$ will correct



other option i don't know...



Any hints/solution will be apprecaited



thanks u






real-analysis







share|cite|improve this question









$endgroup$




let $f(x) = tan ^{-1}x, x in mathbb{R} $ .then



choose the correct statement



$1.$ there exist a polynomial $p(x)$ satisfying $p(x)f'(x) =1$ for all $x$



$2.$$f^{(n)}(0) =0$ for all positive even integer $n$



$3.$ the sequence ${f^{(n)} (0) }$ is unbounded



$4.$$f^{(n)} (0) =0$ for all $n$



My attempt : if $ x =0$ ,then obviously $f(x) =0$ because $tan 0 =0$



from this i see option $2)$ and option $4)$ will correct



other option i don't know...



Any hints/solution will be apprecaited



thanks u






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 19:51









jasminejasmine

1,791418




1,791418








  • 1




    $begingroup$
    For $(2)$ and $(4)$, I think you misunderstood $f^{(n)}(0)$ as $(f(0))^n; f^{(n)}(0)$ means the $n^{th}$ derivative of $f$ at $x=0$.
    $endgroup$
    – Shubham Johri
    Dec 12 '18 at 20:45












  • $begingroup$
    thanks @ShubhamJohri...
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:53














  • 1




    $begingroup$
    For $(2)$ and $(4)$, I think you misunderstood $f^{(n)}(0)$ as $(f(0))^n; f^{(n)}(0)$ means the $n^{th}$ derivative of $f$ at $x=0$.
    $endgroup$
    – Shubham Johri
    Dec 12 '18 at 20:45












  • $begingroup$
    thanks @ShubhamJohri...
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:53








1




1




$begingroup$
For $(2)$ and $(4)$, I think you misunderstood $f^{(n)}(0)$ as $(f(0))^n; f^{(n)}(0)$ means the $n^{th}$ derivative of $f$ at $x=0$.
$endgroup$
– Shubham Johri
Dec 12 '18 at 20:45






$begingroup$
For $(2)$ and $(4)$, I think you misunderstood $f^{(n)}(0)$ as $(f(0))^n; f^{(n)}(0)$ means the $n^{th}$ derivative of $f$ at $x=0$.
$endgroup$
– Shubham Johri
Dec 12 '18 at 20:45














$begingroup$
thanks @ShubhamJohri...
$endgroup$
– jasmine
Dec 12 '18 at 20:53




$begingroup$
thanks @ShubhamJohri...
$endgroup$
– jasmine
Dec 12 '18 at 20:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

Yes to 1, viz. $p=1+x^2$. Yes to 2, because $f$, like $tan x$, is odd and infinitely differentiable. But 4 is false, viz. e.g. $n=1$. As for 3, note that $f=x-x^3/3+x^5/5-cdots$, so the odd derivaties at $0$ are $(-1)^n(2n)!$, so 3 is true.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @J. G im not getting how option 3, 4 false can u elaborate more in detail ???
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:02










  • $begingroup$
    @BarryCipra Sorry; I misread it as restricting to even $n$.
    $endgroup$
    – J.G.
    Dec 12 '18 at 20:05










  • $begingroup$
    THanks u @JG ..
    $endgroup$
    – jasmine
    Dec 12 '18 at 21:00











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Yes to 1, viz. $p=1+x^2$. Yes to 2, because $f$, like $tan x$, is odd and infinitely differentiable. But 4 is false, viz. e.g. $n=1$. As for 3, note that $f=x-x^3/3+x^5/5-cdots$, so the odd derivaties at $0$ are $(-1)^n(2n)!$, so 3 is true.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @J. G im not getting how option 3, 4 false can u elaborate more in detail ???
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:02










  • $begingroup$
    @BarryCipra Sorry; I misread it as restricting to even $n$.
    $endgroup$
    – J.G.
    Dec 12 '18 at 20:05










  • $begingroup$
    THanks u @JG ..
    $endgroup$
    – jasmine
    Dec 12 '18 at 21:00
















2












$begingroup$

Yes to 1, viz. $p=1+x^2$. Yes to 2, because $f$, like $tan x$, is odd and infinitely differentiable. But 4 is false, viz. e.g. $n=1$. As for 3, note that $f=x-x^3/3+x^5/5-cdots$, so the odd derivaties at $0$ are $(-1)^n(2n)!$, so 3 is true.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @J. G im not getting how option 3, 4 false can u elaborate more in detail ???
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:02










  • $begingroup$
    @BarryCipra Sorry; I misread it as restricting to even $n$.
    $endgroup$
    – J.G.
    Dec 12 '18 at 20:05










  • $begingroup$
    THanks u @JG ..
    $endgroup$
    – jasmine
    Dec 12 '18 at 21:00














2












2








2





$begingroup$

Yes to 1, viz. $p=1+x^2$. Yes to 2, because $f$, like $tan x$, is odd and infinitely differentiable. But 4 is false, viz. e.g. $n=1$. As for 3, note that $f=x-x^3/3+x^5/5-cdots$, so the odd derivaties at $0$ are $(-1)^n(2n)!$, so 3 is true.






share|cite|improve this answer











$endgroup$



Yes to 1, viz. $p=1+x^2$. Yes to 2, because $f$, like $tan x$, is odd and infinitely differentiable. But 4 is false, viz. e.g. $n=1$. As for 3, note that $f=x-x^3/3+x^5/5-cdots$, so the odd derivaties at $0$ are $(-1)^n(2n)!$, so 3 is true.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 20:05

























answered Dec 12 '18 at 19:59









J.G.J.G.

28k22844




28k22844












  • $begingroup$
    @J. G im not getting how option 3, 4 false can u elaborate more in detail ???
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:02










  • $begingroup$
    @BarryCipra Sorry; I misread it as restricting to even $n$.
    $endgroup$
    – J.G.
    Dec 12 '18 at 20:05










  • $begingroup$
    THanks u @JG ..
    $endgroup$
    – jasmine
    Dec 12 '18 at 21:00


















  • $begingroup$
    @J. G im not getting how option 3, 4 false can u elaborate more in detail ???
    $endgroup$
    – jasmine
    Dec 12 '18 at 20:02










  • $begingroup$
    @BarryCipra Sorry; I misread it as restricting to even $n$.
    $endgroup$
    – J.G.
    Dec 12 '18 at 20:05










  • $begingroup$
    THanks u @JG ..
    $endgroup$
    – jasmine
    Dec 12 '18 at 21:00
















$begingroup$
@J. G im not getting how option 3, 4 false can u elaborate more in detail ???
$endgroup$
– jasmine
Dec 12 '18 at 20:02




$begingroup$
@J. G im not getting how option 3, 4 false can u elaborate more in detail ???
$endgroup$
– jasmine
Dec 12 '18 at 20:02












$begingroup$
@BarryCipra Sorry; I misread it as restricting to even $n$.
$endgroup$
– J.G.
Dec 12 '18 at 20:05




$begingroup$
@BarryCipra Sorry; I misread it as restricting to even $n$.
$endgroup$
– J.G.
Dec 12 '18 at 20:05












$begingroup$
THanks u @JG ..
$endgroup$
– jasmine
Dec 12 '18 at 21:00




$begingroup$
THanks u @JG ..
$endgroup$
– jasmine
Dec 12 '18 at 21:00


















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