+2 @geometry dash Let P(m,n) denote the assertion that
f(m+n)=f(m)+f(n)−2f(mn+m+n+1)+m2+n2.f(m+n) = f(m) + f(n) - 2f(mn + m + n + 1) + m^2 + n^2.f(m+n)=f(m)+f(n)−2f(mn+m+n+1)+m2+n2.
Start with P(0,0):
f(0+0)=f(0)+f(0)−2f(0+0+0+1)+02+02.f(0 + 0) = f(0) + f(0) - 2f(0 + 0 + 0 + 1) + 0^2 + 0^2.f(0+0)=f(0)+f(0)−2f(0+0+0+1)+02+02.
This simplifies to:
f(0)=2f(0)−2f(1).f(0) = 2f(0) - 2f(1).f(0)=2f(0)−2f(1).
Given that f(1)=0f(1) = 0f(1)=0, we have:
f(0)=2f(0)−0 ⟹ f(0)=2f(0).f(0) = 2f(0) - 0 \implies f(0) = 2f(0).f(0)=2f(0)−0⟹f(0)=2f(0).
This implies f(0)=0.
Next, we apply P(m,0):
f(m+0)=f(m)+f(0)−2f(m⋅0+m+0+1)+m2+02.f(m + 0) = f(m) + f(0) - 2f(m \cdot 0 + m + 0 + 1) + m^2 + 0^2.f(m+0)=f(m)+f(0)−2f(m⋅0+m+0+1)+m2+02.
This simplifies to:
f(m)=f(m)+0−2f(m+1)+m2.f(m) = f(m) + 0 - 2f(m + 1) + m^2.f(m)=f(m)+0−2f(m+1)+m2.
Rearranging gives:
0=−2f(m+1)+m2 ⟹ 2f(m+1)=m2 ⟹ f(m+1)=m22.0 = -2f(m + 1) + m^2 \implies 2f(m + 1) = m^2 \implies f(m + 1) = \frac{m^2}{2}.0=−2f(m+1)+m2⟹2f(m+1)=m2⟹f(m+1)=2m2.
To express f(m), we note that m+1m + 1m+1 can be replaced by nnn, so:
f(n)=(n−1)22.f(n) = \frac{(n-1)^2}{2}.f(n)=2(n−1)2.
Substituting n=m+1n = m + 1n=m+1:
f(m)=(m−1)22.f(m) = \frac{(m-1)^2}{2}.f(m)=2(m−1)2.
We can compute f(m)f(m)f(m) for any nonnegative integer mmm:
For m=0m = 0m=0:
f(0)=(−1)22=12.f(0) = \frac{(-1)^2}{2} = \frac{1}{2}.f(0)=2(−1)2=21.
For m=1m = 1m=1:
f(1)=0(given).f(1) = 0 \quad \text{(given)}.f(1)=0(given).
For m=2m = 2m=2:
f(2)=(2−1)22=12.f(2) = \frac{(2-1)^2}{2} = \frac{1}{2}.f(2)=2(2−1)2=21.
For m=3m = 3m=3:
f(3)=(3−1)22=42=2.f(3) = \frac{(3-1)^2}{2} = \frac{4}{2} = 2.f(3)=2(3−1)2=24=2.
For m=4m = 4m=4:
f(4)=(4−1)22=92.f(4) = \frac{(4-1)^2}{2} = \frac{9}{2}.f(4)=2(4−1)2=29.
We can generalize to obtain:
f(n)=(n−1)22.f(n) = \frac{(n-1)^2}{2}.f(n)=2(n−1)2.
Now we compute f(123)f(123)f(123):
f(123)=(123−1)22=12222=148842=7442.f(123) = \frac{(123 - 1)^2}{2} = \frac{122^2}{2} = \frac{14884}{2} = 7442.f(123)=2(123−1)2=21222=214884=7442.
+2 Here's a structured approach to solving the functional equation:
Given:
f(m+n)=f(m)+f(n)−2f(mn+m+n+1)+m2+n2f(m + n) = f(m) + f(n) - 2f(mn + m + n + 1) + m^2 + n^2f(m+n)=f(m)+f(n)−2f(mn+m+n+1)+m2+n2
for all nonnegative integers m,nm, nm,n, and f(1)=0f(1) = 0f(1)=0.
Base Case: f(1)=0f(1) = 0f(1)=0 is given.
Plugging in Small Values:
Set m=0m = 0m=0, n=0n = 0n=0, and find f(0)f(0)f(0).
Set m=1m = 1m=1, n=0n = 0n=0, and analyze.
Continue deriving a pattern for f(n)f(n)f(n).
Inductive or Substitution Approach: Solve iteratively for f(123)f(123)f(123).
If you're interested in a detailed breakdown of the solution, check out dreieck prozentrechnung for more mathematical tools and solutions.
+2 I just typed out how to do it and then my post was deleted 
I'm just gonna give the summary:
fill in a,b,c,d for the blanks.
Then, find a common denominator for all variables on the right side. At this point it should look like the image below:
8x3+24x2+15x+1(x+1)(x−1)(x)(x+3)=a(x+1)x(x+3)(x+1)(x−1)(x)(x+3)+b(x+1)x(x−1)(x+1)(x−1)(x)(x+3)+c(x+1)(x−1)(x+3)(x+1)(x−1)(x)(x+3)+d(x−1)x(x+3)(x+1)(x−1)(x)(x+3)
You can cancel the denominator on both sides and foil the polynomials on the right side (but do not multiply by a,b,c or d):
8x3+24x2+15x+1=a(x3+4x2+3x)+b(x3+x2−x−1)+c(x3+2x2−3x)+d(x3−2x2+1)
x3 can only be paired with x3, x2 with x2, and so on. This means ax3+bx3+cx3+dx3=8x3, or a+b+c+d=8. We can apply this rule to x3,x2,x, and the constant. So we get:
a+b+c+d=84a+b+2c−3d=243a−b−3c=15−c+d=1
You may use any method to solve this system of equations. The first step is to substitute d=c+1 into every equation. I used matrices to solve from there, but you may want to add/subtract equations depending on your knowledge. You should get the final answer:
| a=469,b=5,c=−149,d=−59 |
p.s: you would probably have more success in getting replies if you typed your equation into the website's built-in LaTeX editor. Just sayin'
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