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Let-f-x-be-a-real-valuedfunction-x-R-Determine-the-following-sums-1-S-1-n-r-1-n-tan-1-rf-x-2-S-2-n-r-1-n-sin-1-rf-x-f-x-1-n-3-S-3-n-r-1-n-cos-1-




Question Number 1437 by 112358 last updated on 03/Aug/15
Let f(x) be a real valuedfunction  ∀x∈R. Determine the following  sums :  (1) S_1 (n)=Σ_(r=1) ^n tan^(−1) (rf(x))  (2) S_2 (n)=Σ_(r=1) ^n sin^(−1) (rf(x))    {∣f(x)∣≤1/n}  (3) S_3 (n)=Σ_(r=1) ^n cos^(−1) (rf(x))      {∣f(x)∣≤1/n}  (4) S_4 (n)=Σ_(r=1) ^n sinh^(−1) (rf(x))  (5) S_5 (n)=Σ_(r=1) ^n cosh^(−1) (rf(x))    {f(x)≥1/n}  (6) S_6 (n)=Σ_(r=1) ^n tanh^(−1) (rf(x))      {∣f(x)∣<1/n}
$${Let}\:{f}\left({x}\right)\:{be}\:{a}\:{real}\:{valuedfunction} \\ $$$$\forall{x}\in\mathbb{R}.\:{Determine}\:{the}\:{following} \\ $$$${sums}\:: \\ $$$$\left(\mathrm{1}\right)\:{S}_{\mathrm{1}} \left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{−\mathrm{1}} \left({rf}\left({x}\right)\right) \\ $$$$\left(\mathrm{2}\right)\:{S}_{\mathrm{2}} \left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}^{−\mathrm{1}} \left({rf}\left({x}\right)\right)\:\:\:\:\left\{\mid{f}\left({x}\right)\mid\leqslant\mathrm{1}/{n}\right\} \\ $$$$\left(\mathrm{3}\right)\:{S}_{\mathrm{3}} \left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}^{−\mathrm{1}} \left({rf}\left({x}\right)\right)\:\:\:\:\:\:\left\{\mid{f}\left({x}\right)\mid\leqslant\mathrm{1}/{n}\right\} \\ $$$$\left(\mathrm{4}\right)\:{S}_{\mathrm{4}} \left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sinh}^{−\mathrm{1}} \left({rf}\left({x}\right)\right) \\ $$$$\left(\mathrm{5}\right)\:{S}_{\mathrm{5}} \left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{cosh}^{−\mathrm{1}} \left({rf}\left({x}\right)\right)\:\:\:\:\left\{{f}\left({x}\right)\geqslant\mathrm{1}/{n}\right\} \\ $$$$\left(\mathrm{6}\right)\:{S}_{\mathrm{6}} \left({n}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{tanh}^{−\mathrm{1}} \left({rf}\left({x}\right)\right)\:\:\:\:\:\:\left\{\mid{f}\left({x}\right)\mid<\mathrm{1}/{n}\right\} \\ $$$$ \\ $$
Commented by prakash jain last updated on 05/Aug/15
Since most of these sums can be written  as addition. The sum is simply product.  Am i missing something here?
$$\mathrm{Since}\:\mathrm{most}\:\mathrm{of}\:\mathrm{these}\:\mathrm{sums}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written} \\ $$$$\mathrm{as}\:\mathrm{addition}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{simply}\:\mathrm{product}. \\ $$$$\mathrm{Am}\:\mathrm{i}\:\mathrm{missing}\:\mathrm{something}\:\mathrm{here}? \\ $$
Commented by 112358 last updated on 03/Aug/15
Let y=sinh^(−1) x  ⇒sinhy=x  ∵ sinhy=((e^y −e^(−y) )/2)⇒((e^y −e^(−y) )/2)=x  e^y −e^(−y) =2x  ×e^y :⇒e^(2y) −2xe^y −1=0  ∴ e^y =((2x±(√(4x^2 +4)))/2)=x±(√(x^2 +1))  ⇒y=ln(x±(√(x^2 +1)))  (√(x^2 +1))>(√x^2 )=x  x−(√(x^2 +1))<0⇒y∈C ⇔ y=ln(x−(√(x^2 +1)))  ∴ for y∈R, y=ln(x+(√(x^2 +1)))    [∵(√(x^2 +1))>−x]  i.e sinh^(−1) x=ln(x+(√(x^2 +1)))  Also, tanh^(−1) x=(1/2)ln(((1+x)/(1−x)))    ∣x∣<1  cosh^(−1) x=ln(x+(√(x^2 −1)))          x≥1
$${Let}\:{y}={sinh}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow{sinhy}={x} \\ $$$$\because\:{sinhy}=\frac{{e}^{{y}} −{e}^{−{y}} }{\mathrm{2}}\Rightarrow\frac{{e}^{{y}} −{e}^{−{y}} }{\mathrm{2}}={x} \\ $$$${e}^{{y}} −{e}^{−{y}} =\mathrm{2}{x} \\ $$$$×{e}^{{y}} :\Rightarrow{e}^{\mathrm{2}{y}} −\mathrm{2}{xe}^{{y}} −\mathrm{1}=\mathrm{0} \\ $$$$\therefore\:{e}^{{y}} =\frac{\mathrm{2}{x}\pm\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}={x}\pm\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{y}={ln}\left({x}\pm\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}>\sqrt{{x}^{\mathrm{2}} }={x} \\ $$$${x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}<\mathrm{0}\Rightarrow{y}\in\mathbb{C}\:\Leftrightarrow\:{y}={ln}\left({x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\therefore\:{for}\:{y}\in\mathbb{R},\:{y}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:\:\:\:\left[\because\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}>−{x}\right] \\ $$$${i}.{e}\:{sinh}^{−\mathrm{1}} {x}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$${Also},\:{tanh}^{−\mathrm{1}} {x}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:\:\:\:\mid{x}\mid<\mathrm{1} \\ $$$${cosh}^{−\mathrm{1}} {x}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\:\:\:\:\:\:\:\:\:{x}\geqslant\mathrm{1} \\ $$
Commented by 112358 last updated on 05/Aug/15
I′m wondering which of these   sums has an explicit formula  for its evaluation and the ways  by which they can be found. This  post is purely consequent of my  curiosity of which series of   inverse trigonometric functions  and inverse hyperbolic functions  have formulae in finding their  sums. I haven′t yet executed a  full scale investigation of these  sums. I′ve only tried obtaining an  answer for S_1 (n) for n>2 but to  no avail. What topic(s) can I   read up on to help me with this  question?
$${I}'{m}\:{wondering}\:{which}\:{of}\:{these}\: \\ $$$${sums}\:{has}\:{an}\:{explicit}\:{formula} \\ $$$${for}\:{its}\:{evaluation}\:{and}\:{the}\:{ways} \\ $$$${by}\:{which}\:{they}\:{can}\:{be}\:{found}.\:{This} \\ $$$${post}\:{is}\:{purely}\:{consequent}\:{of}\:{my} \\ $$$${curiosity}\:{of}\:{which}\:{series}\:{of}\: \\ $$$${inverse}\:{trigonometric}\:{functions} \\ $$$${and}\:{inverse}\:{hyperbolic}\:{functions} \\ $$$${have}\:{formulae}\:{in}\:{finding}\:{their} \\ $$$${sums}.\:{I}\:{haven}'{t}\:{yet}\:{executed}\:{a} \\ $$$${full}\:{scale}\:{investigation}\:{of}\:{these} \\ $$$${sums}.\:{I}'{ve}\:{only}\:{tried}\:{obtaining}\:{an} \\ $$$${answer}\:{for}\:{S}_{\mathrm{1}} \left({n}\right)\:{for}\:{n}>\mathrm{2}\:{but}\:{to} \\ $$$${no}\:{avail}.\:{What}\:{topic}\left({s}\right)\:{can}\:{I}\: \\ $$$${read}\:{up}\:{on}\:{to}\:{help}\:{me}\:{with}\:{this} \\ $$$${question}? \\ $$

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