Help: Functional Equation

If f(x)f(x) is a polynomial in xx satisfying f(x)f(y)+2=f(x)+f(y)+f(xy)f(x) f(y) + 2 = f(x) + f(y) + f(xy) and f(0)=0,f(1)=2f(0) = 0, f'(1) = 2 , find f(x)f(x) .

Note by Akhilesh Prasad
3 years, 8 months ago

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Is the question all right?? Put y=0 in the given functional equation to get: 0+2=f(x)+0+0    (since f(0)=0)0+2=f(x)+0+0~~~~\small{\text{(since f(0)=0)}} f(x)=2\Large f(x)=2 which does not satisfy the condition f'(1)=2 !!

(It might work if indeed it was given f(1)=2 and f'(0)=0)

Rishabh Jain - 3 years, 8 months ago

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I also came up with the same conclusion as you, i.e. f(x)=2f(x)=2, so i thought of using the f(1)=2f'(1)=2. So as we know from the definition of derivatives

f(1)=f(1+h)f(1)hf'(1)=\frac { f(1+h)-f(1) }{ h }\\

f(1)=f(1(1+h))f(1)h\Longrightarrow f'(1)=\frac { f(1(1+h))-f(1) }{ h }

From the functional equation we have, \\

f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))

f(1+h)=f(1)f(1+h)f(1)f(1+h)+2\Longrightarrow f(1+h)=f(1)f(1+h)-f(1)-f(1+h)+2

f1(x)=f(1)f(1+h)f(1)f(1+h)+2f(1)h=f(1)(f(1+h)2)(f(1+h)2)h\therefore \quad f^{ 1 }\left( x \right) =\frac { f(1)f(1+h)-f(1)-f(1+h)+2-f(1) }{ h } =\frac { f(1)(f(1+h)-2)-(f(1+h)-2) }{ h }

f1(x)=(f(1)1)(f(1+h)2)h\Longrightarrow f^{ 1 }\left( x \right) =\frac { (f(1)-1)(f(1+h)-2) }{ h }

2h=(f(1)1)(f(1+h)2)\Longrightarrow 2h=(f(1)-1)(f(1+h)-2)

From here on I am not able to conclude anything useful, so if you would help me lead in the right direction or tell me something that i am doing wrong.

Akhilesh Prasad - 3 years, 8 months ago

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You have to take the limit for the derivative, as h goes to 0.

Ameya Daigavane - 3 years, 8 months ago

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Please if you can post a solution@parv mor

Akhilesh Prasad - 3 years, 8 months ago

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f(1)=2 and f'(0)=0 should be there for the question to be correct

parv mor - 3 years, 8 months ago

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But your suggestion makes the functional equation pretty much useless, as for x=1x=1, and y=cy=c, cdom(f)c\in dom(f) we do not get any relation to solve for f(x)f(x)

Akhilesh Prasad - 3 years, 8 months ago

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Ya.. I also reached the same conclusion...

Rishabh Jain - 3 years, 8 months ago

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Hey, if you see the other post, which is the same question, here,
you'll see we get,
f(x)1=xt f(x) - 1 = x^t
For no value of t, will both the given conditions be satisfied- there is no such function, as a result. Try it out! I think you should check your question again.

Ameya Daigavane - 3 years, 8 months ago

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The question is wrong . I will prove it . Put x,y=0x,y=0 f(0)23f(0)+2=0f(0)^{2}-3f(0)+2=0 This implies f(0)=1,2f(0)=1,2 now put x,y=1x,y=1 this gives f(1)=1,2f(1)=1,2 Differentiating the function partially with respect to xx we get f(x)f(y)=f(x)+yf(xy).......(1)f'(x)f(y)=f'(x)+yf'(xy).......(1) Now put x,y=1x,y=1 in (1)(1) We get f(1)=2f(1)=2 Now put x,y=0x,y=0 in (1)(1) We getf(0)=1..or..f(0)=0f(0)=1..or..f'(0)=0 Now put x=1x=1 in (1) we get 2f(y)2y=f(y)...(2)\frac{2f(y)-2}{y}=f'(y)...(2) Now let f(0)=1f(0)=1 , we have f(1)=2f(1)=2then we observe that f(x)=x+1f(x)=x+1 also in satisfies main functional equation and (2),(1) but f(x)=1f'(x)=1 hence question is wrong . If only main functional equation is given without any condition then f(x)=x+1f(x)=x+1.

Shivam Jadhav - 3 years, 8 months ago

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