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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-

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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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