Show that f: R->R defined as f(x) =4x/(3x+4) is bijective.
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Answer:
Nicholas McConnell, studies Mathematics at Rutgers University (2021)
Answered Apr 16
For any fixed α∈C, we have the Newtonian expansion
(1+x)α=1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+…,
obtained by taking the Taylor series of (1+x)α centered at the origin.
If α is a non negative integer n, observe that the coefficient of xk is (nk) for k⩽n and 0 for k>n (in which case the numerator has a factor of zero). In this special case the series is the polynomial which is the familiar binomial expansion of (1+x)n.
However, if α is not a non negative integer (α could be −1, or 12, or even i), all of the coefficients of the series are nonzero and the Ratio Test shows the radius of convergence to be 1. Here are some examples:
Taking α=−1, 11+x=1−x+x2−x3+x4−x5+…
Taking α=12, 1+x−−−−−√=1+12x−18x2+116x3−5128x4+7256x5−…
Taking α=i, (1+x)i=1+ix−12(1+i)x2+(12+16i)x3−512x4+(13−112i)x5+…
Yet those only converge for |x|<1 (any maybe some |x|=1).
This idea can be utilized in many ways:
If A is a commutative Q-algebra, consider the power series ring A[[x]]. Let Z be the set of power series with constant term 0 and U the set of power series with constant term 1. Then Z is an additive abelian group, and U is a multiplicative abelian group. There’s an interesting fact that for n≠0 in Z, the map f↦fn of U (which is manifestly a group endomorphism) is bijective. Feel free to try to work out why this is so.
Using this one may invert the automorphism f↦fn to get f↦f1/n. Composing automorphisms, we have automorphisms f↦fr on U for any r≠0 in Q.
Moreover, we have an exponential map exp:Z→U sending f↦∑∞n=0fnn!, and a logarithm map log:U→Z sending 1+f↦∑∞n=1(−1)n−1fnn. (Both summations are well-defined because f∈Z and hence each coefficient has finitely many nonzero summands.) It can be shown that these maps are group isomorphisms which are inverses of one another. Moreover, for r∈Q, fr=exp(rlogf) for f∈U.
An easy consequence is that the coefficients of (1+x)r are polynomials in r. Since these coefficients agree with the Newtonian expansion’s polynomial coefficients on the infinitude of nonnegative integers (the distributivity of multiplication over addition expands (1+x)n,n∈N easily), it follows that the polynomials coincide. Hence for rational r, (1+x)r is the Newtonian expansion.
On an unrelated topic, the Newtonian expansion can be used to prove limn→∞(1+xn)n=ex. After all, we have
(1+xn)n=1+n(xn)+n(n−1)2!(xn)2+n(n−1)(n−2)3!(xn)3+…=1+nnx+nnn−1nx22!+nnn−1nn−2nx33!+…=∑k=0∞nnn−1n…n−(k−1)nxkk!=∑k=0∞1(1−1n)…(1−k−1n)xkk!;
As n→∞, since each coefficient keeps k constant, the 1−an terms all converge to 1 and hence vanish as multiplicands, leading the series to ∑∞k=0xkk!=ex.
Now let us cut the digression and answer your question. You wanted to find (1+x)−2. This is simply the Newtonian expansion for α=−2, and in this case α(α−1)…(α−(k−1))k!=(−2)(−3)…(−1−k)k!=(−1)k(k+1). Hence
(1+x)−2=1−2x+3x2−4x3+5x4−6x5+…