Taylor Series and Polynomials


 Alyson Holland
 3 years ago
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1 Taylor Series ad Polyomials Motivatios The purpose of Taylor series is to approimate a fuctio with a polyomial; ot oly we wat to be able to approimate, but we also wat to kow how good the approimatio is. We are already familiar with first order approimatios. The it: si tells us that whe is very close to zero si ca be approimated with. Usig this result we are able to coclude that the itegral: si d coverges if ad oly if is coverget; ad we ca also say that the series: is coverget if ad oly if: d si i is coverget. The first case is a coverget itegral ad the secod case is a diverget series. I both these two cases the precisio provided by the first order approimatio of sie suffices. Let s aalyze the followig eample: si 5 4 d We wat to determie if the improper itegral is coverget or diverget. eample we ca try to use it compariso ad we obtai the followig: si 5 4 ( si 4 4 As i the previous Whe the it is zero, it compariso implies covergece if the itegral of the fuctio we are comparig with is coverget, but it does t imply divergece if the itegral we compare with is diverget. I this specific case the it is zero ad the itegral we are comparig with is: 4 d
2 which is diverget. Limit compariso i this case is icoclusive. We have used that si is approimately zero whe is small, ad because of this approimatio it compariso is icoclusive. Is there a more precise way of approimatig this fuctio? Sice we wat to approimate with a polyomial we ca try to fid out if there is a moomial a b such that: si a b where a is a real umber ad b is a atural umber.this it is a / case, so we ca apply de l Hôpital: Remember ow the fudametal it: si a b cos ab b cos By comparig the two its we ca coclude that b 3 ad a 6, so fially: or equivaletly: si 3 6 si 3 6 We ca ow use this more refied approimatio to solve the improper itegral: si ( si 3 6 We are ow allowed to use it compariso i both the two directios; sice the itegral of is diverget i [, ] we ca coclude that the itegral we are studyig is actually diverget. We could iterate this method ad have better approimatios of si. The fuctio: si approaches to zero as goes to zero. We ca determie how fast by comparig with a moomial c d. We wat to fid c, d such that the followig it is : si c d We ca apply de l Hôpital three times ad fid out that d 5 ad c 5! which meas that si is approimated by: si 3 3! + 5 5! This kid of approach opes two questios: Ca we iterate the method ifiitely may times? What is the precisio of the approimatio at each step? If the aswer to the first questio is positive, we ca write the fuctio si as a power series. We call this kid of power series the Taylor series of si cetered at. I the et sectio we will try to aswer these two questios. ( 6
3 Taylor s theorem It is clear from the eample that we have just studied, that i order to determie a approimatio of a fuctio we eed to calculate some of its derivatives (we have used de l Hôpital!!!. The more the approimatio is precise the higher is the umber of derivatives that we eed to calculate. The followig defiitio comes i hady whe we eed to quatify how may times a fuctio is differetiable: Defiitio.. Let f be a real fuctio defied o a domai D. If the fuctio is cotiuous at every poit i D we say that it belogs to C (D. If the fuctio is differetiable times at each poit of D (ecludig the boudary ad the th derivative is cotiuous, we say that the fuctio is i C (D. If the fuctio ca be differetiated ifiitely may times we say that it is i C (D. Remark.. If a fuctio is i C (D it is also i C k (D for every k. Taylor s theorem says that if a fuctio f( belogs to C + (D ad α D, the the fuctio ca be approimated with a degree polyomial of this kid: P,α ( i f (i (α ( α i i! Theorem.3. Suppose f is a real fuctio i C + ([a, b], the for every poit α (a, b there is a fuctio h ( such that: f( P,α ( + h (( α ad: h ( α Remark.4. We should thik of h (( α as a remaider of our approimatio of f( with the polyomial P,α (. We ca say somethig more precise about the remaider: Theorem.5. I the setup of the previous theorem, for every (a, b there is a poit ζ betwee ad α such that: h (( α f (+ (ζ ( a+ ( +! Patchig the two results together we have that: f( P,α ( + f (+ (ζ ( a+ ( +! Remark.6. Notice that i the previous formula the value of ζ depeds o ; if this were ot the case every fuctio would a polyomial! proof of theorem.3. We ca prove the theorem by iductio startig from. Whe the formula becomes: f( f(α + h ( The fuctio h ( coverges to zero whe goes to α because f( is cotiuous. Assume that the statemet is true for some umber : f( P,α ( + h (( α 3
4 We prove that the statemet is true for +. Maipulatig the previous idetity we have: f( P +,α ( f (+ (α ( +! ( α+ + h(( α P +,α ( + h + (( α + The ew remaider is the fuctio: h + ( h ( α f (+ (α ( +! I order to prove the statemet for +, we have to prove that: or equivaletly: or eve: h +( α h ( α α f (+ (α ( +! f( P,α ( α ( α + f (+ (α ( +! I order to calculate this last it we apply de l Hôpital + times. The ( + th derivative of ( α + is ( +!, the ( + th derivative of P,α ( is zero because the polyomial has degree so the it is precisely f (+ (α (+! sice the ( + th derivative of f( is cotiuous. This completes the iductive argumet. Remark.7. At a closer look, the previous proof is just the method that we have used to approimate si tured ito a iductive argumet. proof of theorem.5. Fi a value of oce ad for all. Deote with g(t the followig fuctio: g(t f(t P,α (t h ( (t α+ α By costructio g(, moreover all the derivatives of g(t at the poit α are zero up to the th oe: g(α g ( (α g ( (α... g ( (α ad for the ( + th derivative we have: g (+ (t f (+ (t h ( ( +! α sice the polyomial P,α (t has degree. I order to prove the statemet, it s eough to prove that there is a poit ζ such that g (+ (ζ. If f(t is i C + ([a, b], the fuctio g(t is also i C + ([a, b]. Sice g(α g( there is a poit betwee α ad such that g ( ( because of the mea value theorem. Sice g ( (α g ( ( there must be a poit betwee α ad such that g ( (. We ca repeat this method up to the th derivative of g ad fid a poit such that g (+ (. The poit is the poit ζ that we are lookig for. 4
5 These two results provide a partial aswer to the two questios that we have formulated at the begiig: if a fuctio ca be differetiated + times we ca approimate it with a degree polyomial ad the error ca be calculated i terms of the ( + th derivative Providig a complete aswer to the first questio is subtler. Theorem.3 might lead us to thik that a C (D fuctio f( is equal to its Taylor series for every poit i the domai D: f( f ( (α ( α ; α D! This is ot true. The statemet that we have prove is true for ay umber, but it does t imply aythig about the possibility of a ifiite series. There are C fuctios that are ot equal to the associated Taylor series ad as a matter of fact they are ot equal to ay power series at all. For eample let F ( be the followig fuctio: F ( { e if if It s ot hard to check that this fuctio is i C (R. I doig the calculatio of the derivatives just remember that the epoetial goes to zero faster tha ay polyomial. Every derivative of F ( is zero at the origi: F ( ( if This meas that the Taylor series of F ( is zero everywhere ad its radius of covergece is +. But the fuctio F ( is zero oly at the origi, so it caot be equal to its Taylor series. This kid of surprisig result is perfectly compatible with theorem.3. The theorem just says that for every the fuctio ca be approimated with zero ad the remaider h ( is just F ( : h F ( ( N 3 Taylor series If a fuctio f( is C + over some iterval [a, b] it makes sese to write the Taylor series cetered at some poit α (a, b: f( f ( (α ( α! Theorem.3 guaratees that the fuctio has a Taylor polyomial of a arbitrarily high degree, but this does t imply aythig about the covergece of the series. The series that we have writte is just some formal writig at this stage, ad it might ot have ay practical purpose. We certaily are i oe of the followig scearios: The series is ot coverget at all The series is coverget i a iterval that is at most [a, b] The series is coverget i some iterval but it does t coverge to f( (the fuctio is ot aalytic 5
6 I order to study the covergece of the series we ca apply the root or ratio test ad determie the radius of covergece, but this is ot eough to determie if the series is actually coverget to the fuctio f(. Ufortuately at the momet we do t have a good criterio to determie if a fuctio is aalytic or ot. The fuctios i the followig list ad their iverses are all aalytic: si, cos, ta, cot, sec, csc, ep, log, ( + α, sih, cosh, tah, coth, sech, csch. 3. The poit at ifiity If a fuctio f( belogs to C + (R it might or might ot have a Taylor series cetered at. Necessary coditio to have a epasio at is: f( L < + + which meas that the fuctio has a horizotal asymptote. If this coditio is ot satisfied the Taylor series would be certaily diverget. This coditio does t suffice though, sice we have already see that a fuctio ca be o aalytic at some poit. Eample 3.. The fuctio arcta has a horizotal asymptote at ± which is ± π. If we wat to try to epad the fuctio at positive ifiity we ca epad the fuctio arcta y at zero istead, sice y goes to + as y approaches +. For this fuctio we ca apply the usual Taylor formula ad we obtai: arcta y π y+ ( + This is a Taylor series i a right eighborhood of zero ad the radius of covergece is oe. I order to calculate the epasio at + of arcta we just make the substitutio y : arcta π + + ( + ( + + ad with the same method at egative ifiity we have: arcta π + + ( + ( + + The fuctio arcta is aalytic both at positive ad egative ifiity. Eample 3.. We ca apply the same method to ( + which has a horizotal asymptote at zero. We epad ( + y at zero ad we obtai: y + + y ( y + ad after the substitutio y we have the epasio at ifiity: + + ( Eample 3.3. The fuctio e has a horizotal asymptote at +. This fuctio is aalytic at every poit i R, but is it aalytic at +? + 6
7 4 Ladau symbols: big O I this sectio we itroduce oe of the Ladau symbols: big O. This otatio provides a coveiet way of writig the remaider whe we approimate a fuctio with a Taylor polyomial. The otatio is particularly helpful whe we eed to study how the error due to the approimatio spreads withi a certai calculatio. For eample suppose that we have a product of two fuctios f(g( ad we approimate each oe of them with a Taylor polyomial up to a certai precisio. We eed to kow what the precisio of the product is, ad the big O otatio provides a very coveiet way of solvig this problem. The defiitio of big O may look kid of obscure at a first sight, but i practice it is very ituitive: Defiitio 4.. Let f(, g( be two real fuctios. The fuctio f( is a big O of g( as approaches to a if there are two positive costats M, δ such that: f( M g( for all satisfyig a < δ The meaig of big O is clarified by the followig propositio: Propositio 4.. If the it: the f( O(g(. f( C < + a g( Remark 4.3. This propositio is ot equivalet to the defiitio!!! But the differece is kid of irrelevat for the scearios that we will cosider i these otes. Usig the propositio we ca calculate the followig eamples: 5 O( at ay poit 3 O( at ay poit si O( as si O( as si O( as si( O( as cos O( as cos O( 3 as ((! O as +, (Stirlig e I order to use the big O otatio it is essetial to uderstad how the O symbol behaves withi a formula; this is a list of its properties: O(f(O(g( O(f(g( ( O(f( + O(g( O( f( + g( ( f( + O(g( O( f( + g( (3 f(o(g( O(f(g( (4 O(cf( O(f( c R, c (5 if a g( < + the O(f(g( O(f( (6 7
8 g( Remark 4.4. Suppose that a f( < + the we ca combie the secod ad the last property to obtai: ( ( O(f( + O(g( O( f( + g( O f( + g( f( O(f( I doig a calculatio with big O we should always try to keep the fuctio withi O as simple as possible by usig the properties that we have listed. For istace, suppose + : ( ( log( ( O + O O 3 + log( ( log( ( O 3 log( + ( log( O 3 At the same time we do t wat to lose precisio: if f( is O( it is also O( i for i but we do t wat to replace O( with a lower epoet. More i geeral: eve if O(f(g( O(f( wheever a g( < +, if the it of g( is zero we lose precisio by replacig O(f(g( with O(f(. Remark 4.5. I order to complete calculatios ivolvig oe or more big O s, it is fudametal to remember that if: a f( the: but we caot say aythig about the followig: O(f( a O(??? a If at the ed of a calculatio we caot get rid of the O( s, this usually meas that we have lost too much precisio i the calculatio. This is the reaso why we should be cocered about losig precisio! Remark 4.6. Somethig similar applies to series ad improper itegrals. We caot say aythig about the covergece of the followig series: but if a certai series: is absolutely coverget, the also: O( a O(a is absolutely coverget eve if we do t kow eactly which sequece O(a is. This follows by compariso. If b O(a, this meas that there is a costat M such that: b M a whe is large eough. The by compariso if a is coverget, also b is coverget. 8
9 This method works with absolute covergece oly ad it caot be used to prove divergece. For eample: ( O < + but we caot say aythig about: sice both ( ad are O(. ( O 5 Remaider of a Taylor polyomial as a big O Propositio 5.. Let f( be a fuctio i C + ([a, b] ad α a poit i (a, b, the Taylor epasio ca be writte i big O otatio: f( P,α ( + O(( α + Proof. As a cosequece of theorem.5 we have: f( P,α ( α ( α + f (+ (α < + ( +! which implies that the remaider is O(( α +. Remark 5.. Eve if O(( α + is a ukow fuctio, it is a fuctio i its ow right, so we ca compose it with aythig we wat. For eample: ad e O(3 e O(6 The same thig applies to series: if we kow that f( is equal to its Taylor series for α < R ad g( is a fuctio such that g( α < R we have: For eample, if < we have: f(g( f ( (α (g( α! + ( we ca compose it with a ad we obtai the epasio: but it holds for < a + a ( a 9
10 6 Applicatios The first applicatio of Taylor polyomials is to the calculatio of its of fuctios ad sequeces. Through it compariso for improper itegrals ad series ad the root/ratio test, Taylor polyomials become a very powerful tool to study the covergece of improper itegrals ad series. Eample 6.. We start from the motivatig eample of the first sectio: si O( + 3 O( + 3 O(+ O(??? We have eded up with a O(, this meas that the precisio does t suffice. We try to take a better approimatio of si : si O(4 + 3 O( O( + 6 Note that i the previous eample we could have used de l Hôpital twice ad get the same result: si + 3 cos + 6 si + 6 Sometimes, however, the method of de l Hôpital will lead to very complicated formulas. The followig eample illustrates this: Eample 6.. Cosider the it The derivative of the umerator aloe is: cos( si( l( +. si( si( l( + + cos( cos( l( + ( l( ad this still teds to as approaches. So to compute the origial it we would have to differetiate the above formula oce more, makig it very easy to do mistakes. Istead, we ca use the big O otatio as follows: so we obtai: cos( si( l( + ( + O( ( l( + + O( 3 l 3 ( + ( + O( ( ( + O( + O( 3 ( + O( ( + O( 3 + O( 3, + O( 3 + O(. Caveat: epadig a compositio of fuctios ca be tricky. I the previous eample we had to calculate si( l( +. I the eample we have decided to epad sie first, ad the log. I geeral it is faster to epad the ier fuctio first, but it s also more delicate. Suppose we wated to epad si(l( + up to O( 4 : si(l( + si( + O( + O( +...
11 o metter how much you epad sie you caot improve O(. To reach the precisio of O( 4 we first eed to epad log up to O( 4 : si(l( + si ( O( O(4 ( O( O(4 6 3 I the epasio of sie we caot stop at the first order, we have to calculate ay order that cotributes moomials up to degree four. I this specific case we eed to calculate the third order but we do t eed to take ay higher order. Eample 6.3. log( + arcta e We ca try to solve the it just by takig Taylor polyomials of legth : log( + ( + O( 3 + O( O( 8 log( + + O( 4 + O( O( 8 + O( 4 + O(( + O( 4 + O( O( 8 O( O( 8 O( + O(4??? The calculatio is icoclusive because we have O( at the umerator. This meas that we have to icrease the precisio of the umerator: log( + ( O( O( O( O( 6 ( O( 6 + O( O( O( O( O( O( + O(4 8 3 I the calculatio we have used the followig epasios: arcta O( 5 e O( 6 log( + + O( O( 8
12 Caveat: If, withi a calculatio, you have both O( ad a moomial a m with m, the moomial ca ad must be discarded: a m + O( O( a m + O( ( + a m O( Eample 6.4. Determie if the followig series is coverget or diverget: ( 8( cos ( si As a preiary step we apply the root test for absolute covergece of the series ad reduce the calculatio to the study of the followig it: 8( cos ( + si ( ( ( + O 4 ( O ( ( ( ( + O O 5 6( ( ( ( + O O 8 ( ( ( ( + O 36 + O 7 < The it is less the, so accordig to the root test the series is absolutely coverget. Strategy: We have approimated cosie at the secod order but sie at the third oe, ad it is ot hard to check that a less precise approimatio of sie produces a icoclusive output. The ituitive reaso why we eed higher precisio for sie is the followig: the term cacels the first order approimatio of sie, but the term oly cacels the zero order approimatio of cosie. The geeral strategy is to epad a fuctio up to the first term that is ot caceled by some other compoet i the formula. This is just a guidelie ad ot a rule: sometimes it might ot be ecessary to epad a fuctio this much, sometimes it might ot suffice! Figurig out at first sight the legth of a Taylor polyomial that suffices i a certai calculatio is a art; till the day you master the art, trial ad error is the oly way! If the polyomial is too short the method is icoclusive; if the polyomial is too log, the calculatio might get epoetially messy! Eample 6.5. Determie if the followig series is coverget/absolutely coverget: ( e 3 + cos si
13 ( + + ( 7 ( ( ( + O O ( + O 4 ( 6 ( O O 6 + O ( ( 5 4! 4 At this poit we might be tempted to use it compariso ad coclude that the series is coverget; but it compariso caot be applied to a alteratig series. Istead of usig it compariso we try to separate the part of the series that coverges but ot absolutely, from the part that coverges absolutely: ( ( ( ( 7 ( 7 ( ( ( + O + O ( 7 + O ( + ( O 3 The first series is coverget ad the secod oe is absolutely coverget because of remark 4.6, ad this implies that their sum is coverget. However the etire series is ot absolutely coverget, if we get rid of ( by takig the absolute value, we ca actually use it compariso ad coclude that the series behaves like which is diverget. Eample 6.6. si( log( + cos( si( + log( + log ( + O( 3 log 3 ( log( + cos( + log( + O( log ( log ( + O( 3 log 3 ( + + ( log( + O( log ( + log ( + O( 3 log 3 ( log ( + O( 3 log 3 ( + + O( log( + O( log( I this calculatio we have used that e log( ad that log( as +. Eample ( +3 3 (4 + (cos 3 ( (si 3 ( 3 cosh( 3
14 + + + ( 3 ( 3 + O(4 3 4 log( + O( 8 log ( + (( + O(4 3 ( O( 5 4 log( ( 9 log( + O( A few remarks about this calculatio: (( O(5 3 3 (O( O( 7 ( 3 + O( + O( 3 the hyperbolic cosie is completely irrelevat sice it coverges to oe. Approimatig fuctios with polyomials is ot a algorithm to solve its, it s just oe of may tools. It s usually a good idea to simplify a it with ay other method before doig ay approimatio. This calculatio would get very log ad messy if we did the algebra completely. But this is ot ecessary sice we are workig with big O. For istace the calculatio of the cube: 3 9 ( O(5 would be very log ad tedious. It is a good idea to calculate the smallest big O first. I this case it is 3 O( 5 O( 7. The smallest big O represets the maimum precisio that we ca achieve. Ay moomial or big O of a moomial with degree higher or equal to 7, does t carry ay iformatio ad it ca be absorbed i O( 7. It turs out that the oly sigificat terms i the cube are: 3 ad 3 ( I the third step we have used: O( 5 + O( 8 log ( O( 5 ( + 3 log ( O( 5 ad the fact that 3 log ( as +. 7 A advaced applicatio: Stirlig s formula I the sectio about sequeces we saw that itegral compariso provides a good estimate for the factorial of a umber: ( + +! e e This estimate suggests that the factorial might behave more or less like e ; but this is ot completely true sice the it: +! + e is ifiity ad ot oe. This meas that the factorial is faster tha e ; we would like to kow how faster it goes to ifiity. Istead of workig with the factorial we will use the gamma fuctio Γ( + + t e t dt, which is related to the factorial i the followig way: Γ( +!. Theorem 7. (Stirlig. +! e π + Γ( + e π 4
15 Proof. We ca thik of Γ( + has a itegral depedig o a parameter, ad we wat to study its asymptotic behvior. I this kid of setup it s usually coveiet to write the itegral i the followig way: + e h(t dt for some fuctio h(t. I the case of Γ we ca do the followig maipulatio: Γ( + + t e t dt + + e log t t dt + e (log u u du where t u. Sice we caot calculate the itegral as it is, we have to approimate the fuctio log u u. This fuctio is icreasig from to, it attais its maimum at ad its decreasig for u >. It goes to both at zero ad ifiity. Sice we wat to approimate the fuctio i a eighborhood of the maimum it is coveiet to shift the fuctio so that the maimum is at the origi: + + e e (log(v+ v dv where u v +. We ca ow epad the fuctio h(v log(v + v at the origi: h(v log(v + v v v + O(v3 v v + O(v3 ad we ca also epad the epoetial: e h(v e v ( + O(v 3 Sice we are itegratig from to + it s ot eough to use a approimatio that is good i a small eighborhood of the origi oly. This approimatio does t work at where the logarithm ad its derivatives are diverget ad it does t work for a arbitrarily large v; however it holds i ay closed iterval [ + ɛ, M] where both ɛ ad M are positive umbers. We ca rewrite the itegral i the followig way: + e ( +ɛ M e (log(v+ v dv + e v ( + O(v 3 dv + +ɛ + M e (log(v+ v dv We will treat the three itegrals separately. I the first itegral we use that log(v + v is icreasig for v < ad we ca choose ɛ such that for v (, + ɛ we have: If > the previous iequality implies: log(v + v (log(v + v + log(v + v + or equivaletly: By compariso we have: (log(v + v ( + log(v + v +ɛ +ɛ e (log(v+ v dv e ( e log(v+ v dv 5
16 Whe goes to ifiity the right had side coverges to zero, ad this implies that the left had side coverges to zero as well. This meas that the first itegral does t give ay cotributio to the calculatio. Now we ca take care of the secod oe. We separate the two summads: M e v +ɛ dv + M +ɛ O(e v v 3 dv I the first of the two itegrals make the substitutio s v ad the itegral becomes: M e s ds +ɛ For what cocers the secod itegral we ca coclude by compariso that: M +ɛ ( O(e v v 3 M dv O e v v 3 dv +ɛ The itegral withi the big O is somethig we ca actually itegrate eplicitly: M +ɛ e v v 3 dv e v ( v + M +ɛ Whe goes to ifiity this fuctio goes to zero so we ca safely igore it. To calculate the third itegral we use the same trick we have used for the first oe. Sice the fuctio log(v + v is decreasig for v > we ca choose M such that log(v + v < for every v > M. This implies that for ay > the followig iequality holds: By compariso we have: + M (log(v + v ( + log(v + v + e (log(v+ v dv e ( e log(v+ v dv The right had side of the iequality certaily coverges to zero whe + which implies that the left had side also coverges to zero. Eve this last piece ca be safely igored i the calculatio. Puttig all the pieces together we have: M + Γ( + e + + e M e s +ɛ e π ds M + π e s +ɛ ds + e s ds π I this proof we have actually show that the asymptotic behavior of the itegral of e h(v is determied oly by the behavior of h(v i eighborhood of the maimum whe h(v log(v+ v. This is actually a very geeral pheomeo, ad the techique that we have used is a baby versio 6
17 of a very geeral ad powerful method used to determie the asymptotic behavior of itegrals depedig o a parameter kow as the statioary phase method or steepest descet method or Laplace method. This method is widely used i the study of Laplace trasforms. This calculatio is also a clear demostratio that the mechaical usage of Taylor polyomials is ot the ultimate techique to calculate its; i this eample we had to combie Taylor polyomials with a very careful study of the behavior of the itegrad i differet domais. This is certaily ot a ucommo sceario i aalysis. 8 Some more eamples Here are more eamples to practice with. Withi brackets you fid the umerical solutio of its, ad the approimatio of itegrads ad summads.. arcta( l( + ( cos( [ ] 3. si( + [ ] l( + + e ( l( l( + e 4 4 cos(si( si( cos( + cos( [ 8] [ 96 5 ] 5. ( l si + [ 6 ], coverget (si d (( l ( 3 + log 4 l( + d [, diverget ] [ l3 ( ], coverget 8. ( si + cos l 3( + [ 5, coverget ] 9. si ( l ( + si si 6 + [ 4, coverget ] 7
18 . +(l 3( arcta(l( + + π [ ] + (l 3 9 Formulas e! + +! for all 3! (7 log( for < (8 + log( + ( for < (9 for < ( ( (! + ( (! ( for ( ( α ( + α for all < ad all real α ( ( si ( +! + 3 3! + 5 for all 5! (3 ( cos (!! + 4 for all 4! (4 B ( 4 ( 4 ta + 3 (! for < π (5 ( E sec for < π (! (6 (! arcsi 4 (! ( + + for (7 arccos π arcsi π arcta arcta ± π (! 4 (! ( + + for (8 ( + + for (9 ( + + for ± radius of covergece ( 8
19 sih cosh tah arsih( artah( + ( +! + 3 3! for all ( 5! (! +! for all ( 4! B 4 (4 (! for < π (3 ( (! 4 (! ( + + for (4 + + for < (5 The symbol B i the formulas stads for the th Beroulli umber, ad E is the th Euler umber. 9
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