Argue why the function f: R R defined by f(x) = x^3 - x^2 + x is invertible. Gue

Question

Argue why the function f: R R defined by f(x) = x^3 - x^2 + x is invertible. Guess a closed form for the sum F_1 + F_3 + ... + F_2n-1, where n > 0 is a natural number and (F_k)_kgreaterthanorequalto are Fibonacci numbers. This closed form may involve Fibonacci numbers. Show by induction that your guess is correct. Find a closed form for the double sum: sigma_1 lessthanorequalto j lessthanorequalto k lessthanorequalto n^3k Attempt to find a closed form for the sum sigma_k = 1^n k^3 by perturbation, only to find a closed form for the following sum sigma_k = 1^n k^2.

Explanation / Answer

1. If y = f (x), then the inverse relation is written as y = f -1 (x). If the inverse is also a function, then we say that the function f is invertible.

A function is invertible if and only if it is bijective, that is surjective (onto) and injective (one-to-one), so your statement is not correct. If you want to determine thatif a function is injective, you assume f ( x ) = f ( y ) and derive x = y .

LET F(x) NOT EQUAL TO F(y)

X3+X2+X NOT EQUAL TO Y3+Y2+Y

HENCE IT IS INVERTIBLE , AS IF X=2 THEN Y= 14, IF Y=2 THEN X= 14

First note that k2=2(k2)+(k1).k2=2(k2)+(k1).
We therefore have

kk2(nk)=k(2(k2)+(k1))(nk)kk2(nk)=k(2(k2)+(k1))(nk)

=2k(k2)(nk)+k(k1)(nk)=2k(k2)(nk)+k(k1)(nk)

=2k(n2)(n2k2)+k(n1)(n1k1)=2k(n2)(n2k2)+k(n1)(n1k1)

=2(n2)2n2+(n1)2n1=2(n2)2n2+(n1)2n1

=((n2)+(n21))2n1=(n+12)2n1.

Here's another approach using calculus,
Binomial theorem
(1+x)n=nk=0(nk)xk(1+x)n=k=0n(nk)xk

Differentiating both sides wrt to x
n(1+x)n1=nk=0k(nk)xk1n(1+x)n1=k=0nk(nk)xk1

Multiplying both sides by x
nx(1+x)n1=nk=0k(nk)xknx(1+x)n1=k=0nk(nk)xk

Differentiating both sides again
nx(n1)(1+x)n2+n(1+x)n1=nx(n1)(1+x)n2+n(1+x)n1=nk=0k2(nk)xk1k=0nk2(nk)xk1

On substituting x=1x=1 we get the required answer as
n(n1)2n2+n2n1=n(n+1)2n2n(n1)2n2+n2n1=n(n+1)2n2

Also note that substituting x=1x=1suggests that the alternating sum vanishes

The same idea can be used to derive the sums of other binomial series.

2. FOR FIBONACCI

There is no general method if the recurrence is allowed to be sufficiently complex--- because, if we are allowed to use just a little bit of mathematical sophistication, it becomes possible to define a recurrence that corresponds to the operation of a Turing machine.

However, informal methods apply just like any other problem-solving exercise. Calculate a few terms, make a guess, try the guess out. Attempt to find a transformation to a better-understood or more manageable formulation of the problem. Solve a simpler or more restricted version of the recurrence first and see if it provokes any ideas.

In your specific case the recurrence seems a bit underspecified. Any constant value of F is a solution. So is any function F(a,b,c)=f(a)F(a,b,c)=f(a), as long as f(a)f(a) is a fixedpoint of ff. For example, F(a,b,c)=amod2F(a,b,c)=amod2 satisfies, since F(F(a,b,c1),b,c1)=F(amod2,b,c1)=amod2F(F(a,b,c1),b,c1)=F(amod2,b,c1)=amod2. Or, as another trivial solution, any function which depends only bb will work.

One way to produce a less-trivial example is by looking at the recurrence as a "program" which does something to aa, if we restrict cc to positive values, or equivalently a composition of functions indexed by cc. We can rewrite it as

Fc=Fc1Fc1Fc=Fc1Fc1

Fc=Fc2Fc2Fc2Fc2Fc=Fc2Fc2Fc2Fc2

Fc=F1F1F1Fc=F1F1F1, 2c2c times

Now, there are potentially many families of functions which satisfy this requirement! They don't necessarily lead to closed forms, but consider F1(a)=a+1F1(a)=a+1. Then F2(a)=a+2F2(a)=a+2, F3(a)=a+4F3(a)=a+4, and in general Fc(a)=a+2c1Fc(a)=a+2c1.

But an equally valid option is F1(a)=F1(a)= the next prime larger than aa. Then F2(a)F2(a) is the second prime larger than aa, F3(a)F3(a) is the fourth prime larger than aa, and so on. (We could incorporate a second parameter into either of these definitions pretty easily.

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