Evaluating $\frac{100x^{100}+98x^{98}+96x^{96}+\cdots+6x^6+4x^4+2x^2}{99x^{99}+97x^{97}+95x^{95}+\cdots+5x^5+3x^3+x}$ for $x=99$ billion

Question

Approximately what is the value of $f$ evaluated at $99$ billion, where \begin{equation*} f(x) = \frac{100x^{100} + 98x^{98} + 96x^{96} + \dotsb + 6x^6+4x^4+2x^2} {99x^{99}+97x^{97}+95x^{95}+\dotsb+5x^5+3x^3+x}\,? \end{equation*}

This is an exam question. The purpose is to note for large $x$, $f(x) \approx \frac{100}{99}x$, but I'm curious how close this approximation is? Certainly the author is assuming here that the roots of the numerator and denominator are nowhere near $99$ billion. I want to know how close the approximation is to the true value. How can we actually compute $f(99\text{ billion})$?

Answer 1

$f(x)$ can be explicitly computed as $$f(x) = \frac{100 x^{104}-102 x^{102}+2 x^2}{99 x^{103}-101 x^{101}+x^3+x}$$

Now just set $x$ to $99$ billions. Do you really want the exact value?

Here it is:

$$\frac{3697296376497267726572248748202039440378893622168736740033868932557908\ 8556256235374714477514507679997593593949109344414606939296420208734101\ 0676469462676931305920929634862131303428234710638975451616180501668137\ 4288041210648062028509873153310821250394340865995372894273089905503376\ 8877371486345647720439561322903521620203089198795236183472711910545136\ 0485678762281740198667032548265228871457866150061692768582269235380656\ 6512432305250121305580895021750348729056509720374375124519470574551353\ 3493736050559453462153562276079705563853545473648174572669828131441147\ 4126180048376162528066653906407647770923634051152194087662753429592288\ 0098248439983536139909364655692178265236513144248146919690193447933318\ 9868625573609810563825955675181203022354204099957893103593399511234818\ 6772024340957132970493789310921637563413775084138110723387088254982549\ 4728503220650500004033605770299423881130357516390331338732272198740764\ 2154944818984755329149224842616398304491753416198555182993205626590511\ 0459734131971351724748364089900007456522783255920000000570594029940000\ 00000003881196000000000000000198000000000}{3697296376497267726572248671992615819468891105982828487383883413009855\ 2181488381381098668226705780600581968853431174565978104075757257866607\ 4364756069501495870305836791988097516926430080835310687276129629930444\ 4219604449498554112850858976747756576236618537672853336397402664057666\ 5160554173616690571170100361117856482955820430138206160215971840461540\ 2056406217065610786471297627735380679938708139332086668426588322864776\ 0032596202231320600218517379293406133529096413226541583445769643883641\ 4423032276643275146844822980726612273151041276936346283704667478957023\ 0000564015016144116466661236899949976550503780387362818666051158264148\ 0697595393445815011603379511000069772132573908438150080613580011750972\ 1242524254838703362313302708329046131950916016371590083156551311747070\ 5253060951478228277223933812396382565790788973790653600703862530321466\ 0655887841062605536700820157543876159661705744757781515226949069413567\ 3589710758739665481224473234674619606515201833230968064856540782028250\ 9684939011083047022498512809000006590361045807000000000480298005000000\ 000000029403000000000000000001}$$

Since you are asking about how good is the approximation of $f(x)$ by $\frac{100}{99} x$ at $99$ billions we can calculate the difference:

$f(99\times 10^9) - \frac{100}{99}99\times10^9= 2.0612\ldots \times 10^{-15}$.

Pretty small.

Answer 2

Using Python big decimals, I get $$100\,000\,000\,000.0000000000000020612203042567...$$

from decimal import Decimal, getcontext

getcontext().prec = 1000000

x = Decimal (99000000000)
n= Decimal('0')
d=Decimal('0')

for i in range(100,0,-2):
  n += i*x**i

for i in range(99,0,-2):
  d += i*x**i

r= n/d

print("r = ", "{:.100f}".format(r) )

Answer 3

If you are patient, make the long division and you will see that $$f(x)=\frac{100 x}{99}\Bigg[1+\frac{1}{4950 x^2}+\frac{101}{490050 x^4}+O\left(\frac{1}{x^6}\right) \Bigg]$$ So, if $x=99\times 10^9$, the second term in brackets is $$\frac 1{4851495\times 10^{19}}=2.06\cdots \times 10^{-26}$$ and the next one would be $2.14\cdots \times 10^{-48}$