Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230


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1 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given point on its grph. For emple, to find n eqution for the tngent line to the grph of y = f() t the point where =, you first find f (). This is the slope of the tngent line, so n eqution for the tngent line in slopeintercept form is y = f () + f(). We cll the function L() = f() + f () the lineriztion of f() t the point where =. Its grph is the tngent line, so this lineriztion, which is polynomil of degree (t most) one, is the best pproimtion of its kind for the function f() if we look t vlues of tht re close to. Of course, one cn t epect line to be very good pproimtion to grph in generl, but one would epect tht grphs of higher degree polynomils (prbols, cubic curves, etc.) could give better pproimtions. Thus, we would like to find higher degree polynomils tht give good pproimtions to functions. We do ectly tht here for few specil functions. We strt with the formul n = n+. This formul (perhps you lredy sw it derived in clss) is vlid for ll nd ll nonnegtive integers n. We cn rewrite this formul s n = n+. Now we see tht the polynomils n (remember n cn be ny nonnegtive integer here) re good pproimtions to the function s long s the bsolute vlue of the difference, tht is, n+, is smll. Clerly this is the cse if itself is reltively smll nd the eponent n + is reltively lrge. For emple, ssuming /2 nd n is t lest, we know tht the polynomil n pproimtes n+ = n+ (/2) /2 with n error no lrger thn = 2.
2 So we hve successfully found polynomils of higher degree tht re good pproimtions to the function, t lest when <. Using the sme formul bove, we cn lso find polynomils tht pproimte ln( + ) nd polynomils tht pproimte rctn(). Replce the in the formul bove with t nd we get t + t 2 t ( ) n t n = + ( )n t n+. + t This formul is vlid for ll t nd ll nonnegtive integers n. Rewrite this s t + t 2 t ( ) n t n + t = ( )n t n+. + t Integrte both sides with respect to t from to (ssuming > ) nd we hve t + t 2 t ( ) n t n + t dt = ( ) n t n+ dt, + t n+ ( ) n t n+ + + ( )n ln( + ) = dt. 4 n + + t Now in order to show tht the polynomils n+ + + ( )n 4 n + re good pproimtions to ln( + ), we need to show tht the bsolute vlue of the differences, ( ) n t n+ dt + t, re smll. Recll from first semester clculus the generl inequlity b). Also recll the inequlity f(t) dt f(t) dt f(t) dt (ssuming g(t) dt if f(t) g(t) for ll t between nd b. We will use both of these inequlities. We will use = nd b =, so we need to ssume for now tht. We will use f(t) = ( )n t n+. Even though is vrible, when integrting with + t respect to t we cn tret it like constnt. Since we re ssuming, we cn sfely ssert tht f(t) n+ + t for ll vlues of t between nd. This mens tht we cn use g(t) = n+ + t.
3 Using both our inequlities with our choices for, b, f(t) nd g(t) long with the eqution bove, we get the following: ( )n n+ (n + ) ln( + ) = ( ) n t n+ dt + t t n+ + t dt n+ + t dt = n+ ln( + ). This is vlid ssuming. If < < we need to use = nd b =. We cn use the sme f(t), but now use g(t) = n+. Doing this results in the inequlity + t ( )n n+ ln( + ) (n + ) n+ ln( + ) when < < (notice tht the logrithm here is negtive). We re lmost done now. From erlier this semester, we know tht ln(y) y whenever y. Now plugging y = + into our inequlity when shows tht ( )n n+ ln( + ) (n + ) n+ ( + ) whenever nd n is ny nonnegtive integer. For the cse where < < we use the fct tht ln(y) = ln(/y) /y whenever < y <. Plugging y = + into our second inequlity when < < shows tht ( )n n+ (n + ) ln( + ) n+ + whenever < <. We hve now found polynomils tht pproimte ln( + ) when < <. More importntly, the qulity of the pproimtions is quntified. For emple, if we re interested in between nd /2, we see tht ( )n n+ (n + ) ln( + ) n+ ( + ) (/2) n+ ( + /2) = 3 2 n+2.
4 Notice tht 3/2 n+2 is very smll when n is lrge, so our pproimtion gets better when n gets lrger. In prticulr, if we wnt to pproimte ln( + ) to within. = 5 for ll between 3 nd /2, we first check tht 2 < 5. This mens tht we cn use n = 8 nd get ln( + ) 8 < 5. When you compre this with using Simpson s rule (for emple) to pproimte ln( + ) = + with n error less thn 5, you cn strt to see the utility of our polynomil pproimtions. We cn lso tie this in with infinite sequences. Plug = /2 into our polynomils of degree t dt, 2, 3,... bove nd we get n infinite sequence of numbers /2, /2 /8, /2 /8 + /24,... tht converges to ln(3/2). Agin, we not only know tht this sequence converges to ln(3/2), we even hve n upper bound on the bsolute vlue of the difference between ln(3/2) nd ny prticulr number in this sequence. We cn use similr process to produce polynomils tht pproimte the inverse tngent function. Strt with our originl formul nd replce with t 2 to get n = n+ t 2 + t 4 t ( ) n t 2n = + ( )n t 2n+2 + t 2. This is vlid for ll vlues of t, since t 2 cn never equl (becuse it s negtive). As we did before, we cn rewrite this eqution s t 2 + t 4 t ( ) n t 2n + t 2 = ( )n t 2n+2 + t 2. Agin s we did before, we integrte both sides of this eqution from to to get t 2 + t 4 t ( ) n t 2n + t 2 dt = ( ) n t 2n+2 + t 2 dt,
5 so fter integrting n+ + + ( )n 7 2n + rctn() = ( ) n t 2n+2 + t 2 dt. Unlike wht we hd with ln( + ), this eqution is vlid for ll vlues of, not just >. In order for the polynomils to pproimte the rctngent, though, we need the integrl on the righthnd side to be smll. This requires to be less thn. We cn estimte in mnner very similr to wht we did before nd end up with the inequlity n+ + + ( )n 7 2n + rctn() 2n+3 rctn() 2n + 3 Estimting the righthnd side is not so difficult; we just use rctn() < π/2. This gives us the pproimtion n+ + + ( )n 7 2n + rctn() π 2n+3 4n + 6. This is even simpler to use thn our pproimtions for ln( + ). Here it s quite pprent tht the righthnd side is very smll if < nd n is big. For emple, suppose we wnt to pproimte rctn() with n error no lrger thn 5 for ll in the intervl [ /2, /2]. We check tht whenever /2. π < 5, so using n = 6 gives us rctn() < 5. We cn gin tie this into sequences. Plugging = / 3 (which is less thn ) into our polynomils of degree, 3, 5,... gives us sequence tht converges to rctn(/ 3) = π/6: 3, 3 9 3, ,... Agin, we not only know tht this sequence converges to π/6, we lso hve n upper bound on the difference between ny of the numbers in this sequence nd π/6.
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