Find f(2) of a Polynomial Function | R and f(2)=5

  • Thread starter utkarshakash
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I think you went too quickly. g(x)=a0+a1x+...+akxk, so g(x)g(y)=a0a0+a0a1xy+a0a2x2y2+...+akayxk+yk. g(xy)=a0+a1xy+...+ak(xy)k = a0+a1xy+...+akxk yk. g(x)g(y) = g(xy) implies a0a0 = a0a0a1 = a1a0a2 = a2...a0ak = akThese are the only possibility for the coefficients of xk. The other coefficients
  • #36
utkarshakash said:
I do not know how to derive a but assuming a=1, I can find k.

You would able to derive it if you used some other properties of f(x) you figured out before.
 
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  • #37
utkarshakash said:
I do not know how to derive a but assuming a=1, I can find k.
In your post 27 you wrote, correctly,
g(x)g(y)=g(xy)
Substitute g(x) = axk in there.
 
  • #38
haruspex said:
In your post 27 you wrote, correctly,

Substitute g(x) = axk in there.

Thanks!

PS-This was the longest thread I have ever started.
 

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