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functional equations

Mathematics
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\[\color{brown}{f(x)+(x+1)^3=2f(x+1)}\]
Find \(\color{#C00}{f(10)}\)

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Other answers:

lol thanks for posting the full question @ParthKohli
\[f(-1) = 2f(0)\]
f(0)+1=2f(1)
wat to do next i 3 variables 2 equations??
\[f(0) = 2f(1) - 1 \\ f(1) = 2f(2) - 8\]So\[f(0) = 2(2f(2) - 8) = 4f(2) - 16\]
So hard...
\[f(0)=4(f(2)-8)-1=4f(2)-16-1=4f(2)-17\]
is f always of the from\[ax^2+bx+c\]
Ah my bad, yes.
No, it's a polynomial. It can be in that form, but we're not sure.
so far we see \[f(10)+11^3=2f(11)\] \[f(9)+10^3=2f(10)\] ---------------------------------------------- \[f(10)=f(9)-2f(11)+11^3-10^3\]
I think that it must be in the form \(ax^3 + bx^2 + cx + d\)
\[f(x-1)+x^3=2f(x)\]
we need a relation between f(9) and f(11)
\[f(9) = 2f(10) - 1000\]\[f(10) = 2f(11) - 1331\]\[\iff f(9) = 2(f(11) - 1331 ) - 1000\]
\[f(9) = 2f(11) - 2662 - 1000\]
That's the relation.
isnt this the solution then
since \[f(9)-2f(11)=3662\] \[f(10)=3662+11^3-10^3\]
OMG!
But it's an integer between 0 and 999.
okay so we found a relationship between numbers 2 units away from each other x-1 and x+1 \[\color{blue}{f(x-1)+x^3=2f(x).......... f(x)+(1+x)^3=2f(1+x)}\] \[f(x-1)+x^3=4f(x+1)+2(x+1)^3\]
\[f(9)-4f(11)=2(11)^3-10^3\]
earlier we used 2 not 4
Ahhh.
Let f(x)=a+bx+cx^2+dx^3 NOTE: f(x) must be a third degree polynomial
Now, f(x)+(x+1)^3 =2f(x+1)
\[\huge f(10)=f(9)-2f(11)+11^3-10^3=???\]
use it to find a,b,c and d
I see, yeah!
u will get a=-13 b=9 c=-3 d=1
Equating coefficients
yep
Yay
now, once u know f(x) u can find f(10)
help me i dont get where you get the coeeffients from
@Jonask a technique.\[ax + by = 2x + 3y\]means\[a=2,b=3\]
oh great
f(x)+(x+1)^3 =2f(x+1) a+bx+cx^2+dx^3 +x^3+3x^2+3x+1 = 2{a+b(x+1) +c(x+1)^2 +d(x+1)^3
can u continue from here?
\[f(10) = 10^3 - 3(10)^2 +9(10) -13 = 1000 - 300 + 90 - 13 \\ = 777 \]
thats correct
\[\huge \color{green}{THANKS}\]
welcome

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