# actual evaluation – Function satisfying \$f^{-1} =f’\$ Answer

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actual evaluation – Function satisfying \$f^{-1} =f’\$

This query is extra difficult than it appears

If \$f^{-1}(x)=f'(x)\$ then \$f(f^{-1}(x))=f(f'(x))\$ or \$x=f(f'(x))\$

First by-product is giving

\$displaystyle 1=f'(f'(x))f”(x)\$

Now supplant \$y=f'(x)\$

\$displaystyle y(y(x))y'(x)=1\$

Instead of the above we’ll decipher a extra common:

\$displaystyle g(y(x))y'(x)=1\$

This one has an answer expressed as

\$displaystyle int_{1}^{y(x)}g(r) mathrm{d} r = x+c\$

Now since we would like \$y(x)=g(x)\$ we’ve got

\$displaystyle int_{1}^{y(x)}y(r) mathrm{d} r = x+c\$

\$int_{1}^{y(x)}y(r) mathrm{d} r\$ is definitely a inquisitive operator which we’ll delineate as \$mathfrak{R}(y(x))_{1}\$

Now this operator has its inverse and that is what we’re on the lookout for. For a duty \$h(x)\$ we wish to know \$s(x)\$ in order that

\$displaystyle mathfrak{R}(s(x))=h(x)\$

We will convene \$mathfrak{R}^{-1}()\$ the integral root.

As a fresh operator we necessity to gape it a miniature bit. We can create a desk for some recognized features first

\$commence{matrix} h(x) & mathfrak{R}^{-1}(h(x))_{1} & frac{1}{2}(x^2-1) & x frac{1}{3}(x^6-1) & x^2 1 – ln(x)(1 – ln(ln(x))) & ln(x) e^{e^x}-e & e^x -ln(x) & frac{1}{x} cos(1)-cos(error(x)) & error(x) frac{2}{3}(x^{frac{3}{4}}-1) & sqrt{x} aim{matrix}\$

We don’t ignore ceaseless time period since integral root may be very delicate operator and we outline in common integral root with abject \$b\$ as:

\$mathfrak{R}(h(x))_{b}=int_{b}^{f(x)}f(r) mathrm{d} r\$

We will revert to the consequence of abject later however for now we are saying that the duty doesn’t must breathe outlined in every single place and abject can ameliorate about it. So, all we necessity to search out the integral root of \$h(x)=x\$.

The operator resembles a regular by-product/integration besides that it’s extending all of it, it’s giving systematize of a quicker outcome.

It is a abysmal query if this operator has exclusive values. (We will intimate one thing into that course however general it is a superb matter for some semester labor.)

Notice that the operator may be very delicate to the ceaseless worth and this one can not breathe ignored.

Since powers of \$x\$ seems from the powers of \$x\$, we might attempt the answer in probably the most common figure as \$g(x)=ax^b+f(x)\$ nevertheless this may persuade us very snappily that for any answer outlined in every single place \$f(x)=0\$, since such options are very restrictive and it isn’t workable to push constants the route we might love to, plane including \$ax^b+c\$ would create an answer that’s ineffective. The route of coping with ceaseless time period is by altering the abject in common.

So, if we prohibit ourselves to world world we’ve got

\$displaystyle mathfrak{R}(ax^b)_{1}=frac{a^{b+2}x^{b(b+1)}}{b+1}-frac{a}{b+1}\$

Since we would like

\$displaystyle mathfrak{R}(ax^b)_{1}= x+c\$

that requires \$b(b+1)=1\$ making the outcome

\$ba^{b+2}x-ab\$

This requires \$ba^{b+2}=1\$ as properly which makes \$a=(b+1)^{frac{1}{b+2}}\$

(Notice that these will not be two completely different integral roots of the identical duty.)

\$displaystyle f(x)=frac{a}{b+1}x^{b+1}=(b+1)^{-frac{1}{b+1}}x^{b+1}\$

the place as we’ve got talked about \$b(b+1)=1\$

The intuition we go for this operator is that the duty in query has by-product thus it’s assumed that it’s a properly behaving duty.

Let us show that these two are the options

Derivative

\$displaystyle f'(x)=(b+1)^{-frac{1}{b+1}}(b+1)x^{b}\$

\$displaystyle f'(x)=(b+1)^{-frac{1}{b+1}+1}x^{b}\$

\$displaystyle f'(x)=(b+1)^{frac{b+1-1}{b+1}}x^{b}\$

\$displaystyle f'(x)=(b+1)^{frac{b}{b+1}}x^{b}\$

Inverse

\$displaystyle x=(b+1)^{-frac{1}{b+1}}h(x)^{b+1}\$

\$displaystyle (b+1)^{frac{1}{b+1}}x=h(x)^{b+1}\$

\$displaystyle (b+1)^{frac{1}{(b+1)(b+1)}}x^{frac{1}{b+1}}=h(x)\$

\$displaystyle f^{-1}=h(x)=(b+1)^{frac{b}{b+1}}x^{b}\$

Finally allow us to employ golden ratio to search out what the options we’re speaking about

\$displaystyle b_{1}=-phi\$,\$ b_{2}=phi-1\$

making the primary answer really complicated and the second actual:

\$displaystyle (-phi+1)^{-frac{1}{-phi+1}}x^{-phi+1}=(-phi)^{-phi}x^{-frac{1}{phi}}\$

\$displaystyle phi^{-frac{1}{phi}}x^{phi}\$

The two options we’ve got are pushed by differing a ceaseless solely, in any other case the integral root has the exclusive answer simply love regular root has. It is workable that there are answers that aren’t analytical or outlined in every single place, in any other case the 2 given options are the one good world options.

Apart from these world options that don’t change if we alter the integral root abject, there’s an preference of getting a neighborhood answer. For this we prolong the duty analytically sooner or later \$x_{0}\$ and question that the duty behaves as \$x\$ concerning the integral root at that time.

Then nearby \$x_{0}\$ that duty will behave as required.

\$displaystyle f(x)=sumlimits_{n=0}^{+infty}c_{n}(x-x_{0})^n \$

\$displaystyle mathfrak{R}(sumlimits_{n=0}^{+infty}c_{n}(x-x_{0})^n)_{x_{0}} = x+c\$

Notice that the change of abject is essential, albeit purely technical for this specific job, as we’ve got assumed that we all know this duty solely round \$x_{0}\$

Then we necessity to have:

\$displaystyle sumlimits_{n=0}^{+infty}frac{c_{n}}{n+1}(sumlimits_{m=0}^{+infty}c_{m}(x-x_{0})^m-x_{0})^{n+1}=x+f(x_{0})+c\$

Obviously we will match all coefficients because the remaining coefficient by \$x\$ ought to breathe \$1\$ and all larger powers \$0\$. Solving all coefficient for \$c_{ok}\$ will give native options for every \$x_{0}\$. We have already got one such non trifling answer, however there might breathe greater than that world one.

However, recognize that the integral root abject doesn’t influence the respond, because the abject offers with ceaseless time period solely, and ceaseless time period is capricious. Instead of \$1\$ for our duty we might take any ceaseless.

This is to say that our native answer once more has the identical decision which we’ve got create once we began from \$1\$.

Unless there’s some inquisitive and undiscovered route the integral root can have greater than two given options, one actual and one other complicated, every native answer is the same as both of the 2 create world options.

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