If-f-x-is-a-polynomial-function-satisfying-f-x-f-1-x-f-x-f-1-x-x-R-0-and-f-3-28-then-f-4-is-equal-to-

Question Number 17093 by 1234Hello last updated on 02/Jul/17

If f (x) is a polynomial function

satisfying f (x) . f (\frac{1}{x}) = f (x) + f (\frac{1}{x});

x \in R - {0} and f (3) = 28, then f (4) is

equal to

Commented by prakash jain last updated on 30/Jun/17

f (x) = \frac{f (\frac{1}{x})}{f (\frac{1}{x}) - 1}

f (x) = \frac{x - 1}{x}

f (\frac{1}{x}) = 1 - x

f (x) + f (\frac{1}{x}) = \frac{x - 1}{x} + 1 - x = \frac{- 1 + 2 x - x^{2}}{2}

= \frac{- {(1 - x)}^{2}}{x}

f (x) (\frac{1}{x}) = \frac{- {(1 - x)}^{2}}{x}

f (x) = \frac{x - 1}{x} s a t i s f i e s

f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x})

f (3) = \frac{2}{3}

f (4) = \frac{3}{4}

f (3) = 28 ?

Commented by mrW1 last updated on 01/Jul/17

I think functions like this

f (x) = 1 + x^{k} or 1 - x^{k}, k \in R

fulfill the condiction :

f (\frac{1}{x}) = 1 \pm x^{- k}

f (x) f (\frac{1}{x}) = (1 \pm x^{k}) (1 \pm x^{- k}) = 1 \pm x^{k} + 1 \pm x^{- k}

= f (x) + f (\frac{1}{x})

for f (3) = 28

1 \pm 3^{k} = 28 ?

\Rightarrow 1 + 3^{3} = 28

\Rightarrow f (x) = 1 + x^{3}

\Rightarrow f (4) = 1 + 4^{3} = 65

Commented by mrW1 last updated on 01/Jul/17

I started with f (x) = a + bx and got

a = 1 and b = \pm 1 and then I replaced x

with x^{k} without change the property

of the function . I am not sure if there

are other functions besides 1 \pm x^{k} .

if no, then the answer for the question is

unique .

e . g . if f (3) = - 20 then

1 - 3^{k} = - 20 \Rightarrow k = \frac{\ln 21}{\ln 3} \approx 2.77

\Rightarrow f (4) = 1 - 4^{k} = - 45.607 .

if f (3) = 6 \Rightarrow 1 + 3^{k} = 6 \Rightarrow k = \frac{\ln 5}{\ln 3}

\Rightarrow f (4) = 1 + 4^{k} = 8.62

Commented by Tinkutara last updated on 01/Jul/17

Thanks Sirs!

Commented by prakash jain last updated on 01/Jul/17

i t s a t i s f i e s g (x) g (\frac{1}{x}) = 1

and as you described

f (x) = 1 \pm g (x) is valid

solution .

coming back to what is asked in

question .

Try an alternative f (x)

g (x) = \frac{10 x - 3}{10 - 3 x}

f (x) = \frac{7 x + 7}{10 - 3 x} \Rightarrow f (3) = 28

f (\frac{1}{x}) = \frac{7 + 7 x}{10 x - 3}

f (x) + f (\frac{1}{x}) = \frac{7 x + 7}{10 - 3 x} + \frac{7 + 7 x}{10 x - 3}

= \frac{(7 x + 7) (10 x - 3 + 10 - 3 x)}{(10 - 3 x) (10 x - 3)}

= \frac{(7 x + 7) (7 x + 7)}{(10 - 3 x) (10 x - 3)} = f (x) f (\frac{1}{x})

f (4) = \frac{7 \times 4 + 7}{10 - 3 \times 4} = - \frac{35}{2}

Commented by prakash jain last updated on 01/Jul/17

Thanks, mrW 1!

Commented by mrW1 last updated on 01/Jul/17

Do you have any ideas if there are other

types of functions with this property ?

I think the function should be of the

form f (x) = 1 \pm g (x) then

f (\frac{1}{x}) = 1 \pm g (\frac{1}{x})

f (x) f (\frac{1}{x}) = [1 \pm g (x)] [1 \pm g (\frac{1}{x})]

= 1 \pm g (x) + g (x) g (\frac{1}{x}) \pm g (\frac{1}{x})

= f (x) + f (\frac{1}{x}) if g (x) g (\frac{1}{x}) = 1

g (x) = x^{k} fulfills this condition

but are there any other function types

for g (x) ?

Commented by prakash jain last updated on 01/Jul/17

g (x) = \frac{{a x}^{2} + b x + c}{{c x}^{2} + b x + a}

g (\frac{1}{x}) = \frac{\frac{a}{x^{2}} + \frac{b}{x} + c}{\frac{c}{x^{2}} + \frac{b}{x} + a} = \frac{a + b x + {c x}^{2}}{c + b x + {a x}^{2}}

g (x) g (\frac{1}{x}) = 1

Commented by mrW1 last updated on 01/Jul/17

{that}^{'} s great!

can we generase that

g (x) = \frac{\underset{k = 0}{\sum^{n}} a_{k} x^{k}}{\underset{k = 0}{\sum^{n}} a_{n - k} x^{k}}

fulfills the condition ?

Commented by mrW1 last updated on 01/Jul/17

that means for f (3) = 28 there is no

unique solution for f (4) .

thank you for showing me the

right thinking direction!

Commented by 1234Hello last updated on 02/Jul/17

Sir, I changed the question . Here

x \in R - {0} but in your chosen f (x)

= \frac{7 x + 7}{10 - 3 x}, x \in R - {\frac{10}{3}} . Therefore

f (x) cannot be this . But it can be when

x \in R - {0} .

Commented by prakash jain last updated on 02/Jul/17

g (x) = (\frac{{a x}^{2} + b x + 1}{x^{2} + b x + a})

9 a + 3 b + 1 = 9 \times 27 + 81 b + 27 a

18 a + 78 b = - 242

9 a + 39 b = - 121

a = \frac{- (121 + 39 b)}{9}

g (x) = \frac{\frac{230}{9} x^{2} - 9 x + 1}{x^{2} - 9 x + \frac{230}{9}}

g (3) = \frac{230 - 27 + 1}{9 - 27 + \frac{230}{9}} = \frac{9 (230 - 27 + 1)}{(81 - 243 + 230)}

= \frac{9 (204)}{68} = 27

g (x) g (\frac{1}{x}) = 1

f (3) = 28

f (4) = 1 + g (4) = 1 + \frac{230 \cdot 4^{2} - 81 \cdot 4 + 9}{9 \cdot 4^{2} - 81 \cdot 4 + 230}

= 1 + \frac{673}{10} = \frac{683}{10}

- - - - - -

9 x^{2} - 81 x + 230

b^{2} - 4 a c = 81^{2} - 4 \times 9 \times 230 < 0

so f (x) is defined .

if you want to create a function

which is undefined at x = 0

choose g (x) as

\frac{{a x}^{2} + b x + 1}{x^{2} + b x + a} \times \frac{1}{x}

g (3) = 27

\frac{9 a + 3 b + 1}{9 + 3 b + a} = 81

y o u g e t a n e q u a t i o n b e t w e e n a a n d

b .

c h o o s e a v a l u e f o r b w h i c h g i v e s

b^{2} - 4 a c < 0 .

Commented by mrW1 last updated on 02/Jul/17

since there are infinite possible functions :

f (x) = 1 \pm \frac{\underset{k = 0}{\sum^{n}} a_{k} x^{k}}{\underset{k = 0}{\sum^{n}} a_{n - k} x^{k}}

with n \in N and a_{0}, a_{1}, \dots, a_{n} \in R

there is no unique solution for

(n, a_{0}, a_{1}, \dots, a_{n}) from the only one equation

f (3) = 1 \pm \frac{\underset{k = 0}{\sum^{n}} a_{k} 3^{k}}{\underset{k = 0}{\sum^{n}} a_{n - k} 3^{k}} = 28

therefore there is also no unique

value for f (4) :

f (4) = 1 \pm \frac{\underset{k = 0}{\sum^{n}} a_{k} 4^{k}}{\underset{k = 0}{\sum^{n}} a_{n - k} 4^{k}}

Commented by prakash jain last updated on 02/Jul/17

Ok . I missed some part in the

question . The question is

specifically asking for polynomial

function .

f (x) = 1 + x^{3} is only correct answer .

Commented by mrW1 last updated on 02/Jul/17

to 1234 hello :

I think the original question was much

more interesting and challenging

than now . Even though there was no

unique answer but one may think in

all directions . I see it as a great gain

from your original question that we

know now that the requested

function which satisfies the functional

equation could and must be of the form

f (x) = 1 \pm \frac{\underset{k = 0}{\sum^{n}} a_{k} x^{k}}{\underset{k = 0}{\sum^{n}} a_{n - k} x^{k}} .

We {hadn}^{'} t lernt so much if you had

said from very begining that f (x)

is a polynominal function, because

that would have restricted the direction

for our thinking .

Commented by 1234Hello last updated on 03/Jul/17

But this was the original question! I

was in a hurry so I first posted that

version . Thanks all for your efforts!