Jonask
functional equations
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Mertsj
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love em or leave em
Jonask
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\[\color{brown}{f(x)+(x+1)^3=2f(x+1)}\]
ParthKohli
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Find \(\color{#C00}{f(10)}\)
Jonask
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lol thanks for posting the full question @ParthKohli
ParthKohli
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\[f(-1) = 2f(0)\]
Jonask
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f(0)+1=2f(1)
Jonask
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wat to do next i 3 variables 2 equations??
ParthKohli
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\[f(0) = 2f(1) - 1 \\ f(1) = 2f(2) - 8\]So\[f(0) = 2(2f(2) - 8) = 4f(2) - 16\]
ParthKohli
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So hard...
Jonask
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\[f(0)=4(f(2)-8)-1=4f(2)-16-1=4f(2)-17\]
Jonask
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is f always of the from\[ax^2+bx+c\]
ParthKohli
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Ah my bad, yes.
ParthKohli
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No, it's a polynomial. It can be in that form, but we're not sure.
Jonask
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so far we see \[f(10)+11^3=2f(11)\]
\[f(9)+10^3=2f(10)\]
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\[f(10)=f(9)-2f(11)+11^3-10^3\]
ParthKohli
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I think that it must be in the form \(ax^3 + bx^2 + cx + d\)
Jonask
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\[f(x-1)+x^3=2f(x)\]
Jonask
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we need a relation between f(9) and f(11)
ParthKohli
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\[f(9) = 2f(10) - 1000\]\[f(10) = 2f(11) - 1331\]\[\iff f(9) = 2(f(11) - 1331 ) - 1000\]
ParthKohli
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\[f(9) = 2f(11) - 2662 - 1000\]
ParthKohli
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That's the relation.
Jonask
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isnt this the solution then
Jonask
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since
\[f(9)-2f(11)=3662\]
\[f(10)=3662+11^3-10^3\]
ParthKohli
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OMG!
ParthKohli
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But it's an integer between 0 and 999.
Jonask
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okay so we found a relationship between numbers 2 units away from each other x-1 and x+1
\[\color{blue}{f(x-1)+x^3=2f(x).......... f(x)+(1+x)^3=2f(1+x)}\]
\[f(x-1)+x^3=4f(x+1)+2(x+1)^3\]
Jonask
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\[f(9)-4f(11)=2(11)^3-10^3\]
Jonask
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earlier we used 2 not 4
ParthKohli
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Ahhh.
ParthKohli
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@Hero
sauravshakya
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Let f(x)=a+bx+cx^2+dx^3
NOTE: f(x) must be a third degree polynomial
sauravshakya
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Now, f(x)+(x+1)^3 =2f(x+1)
Jonask
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\[\huge f(10)=f(9)-2f(11)+11^3-10^3=???\]
sauravshakya
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use it to find a,b,c and d
ParthKohli
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I see, yeah!
sauravshakya
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u will get
a=-13
b=9
c=-3
d=1
ParthKohli
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Equating coefficients
sauravshakya
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yep
ParthKohli
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Yay
sauravshakya
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now, once u know f(x)
u can find f(10)
Jonask
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help me i dont get where you get the coeeffients from
ParthKohli
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@Jonask a technique.\[ax + by = 2x + 3y\]means\[a=2,b=3\]
Jonask
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oh great
sauravshakya
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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
sauravshakya
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can u continue from here?
ParthKohli
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\[f(10) = 10^3 - 3(10)^2 +9(10) -13 = 1000 - 300 + 90 - 13 \\ = 777 \]
sauravshakya
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thats correct
Jonask
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\[\huge \color{green}{THANKS}\]
sauravshakya
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welcome