Whether there are solutions beyond the polinomial function is the most interesting part of the problem. I'm a little sad that it's not covered in the video.
I have a nother idea, Let´s suppose fx=ax+b Then substitute the equation and you can get a and b values ,a=1/b=-3 And after you can find any function value , But question is f(15) Further fx=x-3 Therefore f15=15-3=12 💀
If we use x = 0, we will be able to cancel a lot of things, because they will be 0 thanks to the terms including a regular x. 0 * f(0 + 1) + f(2 * 0 + 3) = 0^2 ---> f(3) = 0. Now that we know that f(3) = 0, and we used x in such a way that the second term became f(3), now we can choose x in such a way that the first term includes f(3). This means that x + 1 = 3, so x = 3 - 1 = 2. So we will fill in x = 2 everywhere: 2 * f(2 + 1) + f(2 * 2 + 3) = 2^2. This gives 2 * f(3) + f(7) = 4. f(3) = 0, so we get f(7) = 4. Now we can repeat this process to get f(7) in the first term: x + 1 = 7, so x = 7 - 1 = 6. Fill in x = 6: 6 * f(6 + 1) + f(2 * 6 + 3) = 6^2 --> 6 * f(7) + f(15) = 36. f(7) = 4, so 6 * 4 + f(7) = 36 --> 24 + f(15) = 36 --> f(15) = 36 - 24 = 12. This was the first thing that stood out to me to do, and it got me to f(15), because it seemed somewhat obvious that it would.
I solved f(15) in the second method, but I wondered if we could determine a closed form solution for f(x). The way you derived f(x) = x - 3 is very clever! Great video!
@FranciszekKlyk Good point. Assuming f(1) is finite, are x = -1 and x = (2^k - 1) for integer k > 1 the only values of x for which we can unambiguously determine f(x) without further boundary conditions?
problem x f(x+1) + f(2x+3) = x² Assume f(x) is a polynomial. Solve by a method of unknown coefficients. Let f(x) = a x² + b x + c Replace in the equation. x [a (x + 1)² + b(x + 1)+ c] + a (2x+3)² + b (2x +3) + c = x² Expand. a x³ + (6a + b - 1) x² + (7a + 3b + c) x + 9a + 3b + c = 0 Solve coefficient system of equations. a = 0 b - 1 = 0 3 b + c = 0 b = 1 c = -3 f(x) = x - 3 f(15) = 12 answer f(15) = 12
I think you should show why you choose certain values of x rather than just saying if x=2 then ... but why 2. my method ... xf(x+1) +f(2x+3)= x^2 finding f(13) we note the recursive nature and hope that using reducing values of the argument of f will give us a solution. f(2x+3)= x^2-xf(x+1) choose x to.get f(15) on.lhs x=6 f(15)=36-6f(7) now choose x to get f7) on the lhs x=2 f(7)=4-2f(3) now choose x to get f(3) on lhs x=0 f(3) =0 substitue back f(7)=4 -0 f(13)=36-6f(7) f(13)=36-24 =12 in thia way we dont just guess values of z we calculate them. if you didn't spot that another approach would be to try x=0,1,2,3,... and see what you get another approach is to change the dummy variable(x) eg choose a=2x+3 or a=x+1. doesnt gain anything here. but cant assume linear
6*f(7) + f(15) = 36
2*f(3) + f(7) = 4
0*f(1) + f(3) = 0
f(3) = 0
f(7) = 4
f(15) = 36 - 24 = 12
Ppp
Glvpb
😊
brilliant
Whether there are solutions beyond the polinomial function is the most interesting part of the problem. I'm a little sad that it's not covered in the video.
Sorry. I don't know how to do it
Thanks keep going I always watch you your channel is stunning 🎉🎉🎉🎉
Much appreciated!
@SyberMath oh you are so kind teacher
Nice!
Thanks!
I have a nother idea,
Let´s suppose fx=ax+b
Then substitute the equation and you can get a and b values ,a=1/b=-3
And after you can find any function value ,
But question is f(15)
Further fx=x-3
Therefore f15=15-3=12 💀
You should be able to solve it in memory by taking x=0 then x=2 then x=6
x=0..f(3)=0...x=2..2f(3)+f(7)=4=>f(7)=4..x=6...6f(7)+f(15)=36=>f(15)=36-24=12
without watching:
let x = 0, then f(3) = 0
x = 2, 2f(3) + f(7) = f(7) = 4
x = 6, 6f(7) + f(15) = 24 + f(15) = 36
therefore f(15) = 12
If we use x = 0, we will be able to cancel a lot of things, because they will be 0 thanks to the terms including a regular x.
0 * f(0 + 1) + f(2 * 0 + 3) = 0^2 ---> f(3) = 0.
Now that we know that f(3) = 0, and we used x in such a way that the second term became f(3), now we can choose x in such a way that the first term includes f(3). This means that x + 1 = 3, so x = 3 - 1 = 2. So we will fill in x = 2 everywhere:
2 * f(2 + 1) + f(2 * 2 + 3) = 2^2. This gives 2 * f(3) + f(7) = 4. f(3) = 0, so we get f(7) = 4.
Now we can repeat this process to get f(7) in the first term: x + 1 = 7, so x = 7 - 1 = 6. Fill in x = 6:
6 * f(6 + 1) + f(2 * 6 + 3) = 6^2 --> 6 * f(7) + f(15) = 36. f(7) = 4, so 6 * 4 + f(7) = 36 --> 24 + f(15) = 36 --> f(15) = 36 - 24 = 12.
This was the first thing that stood out to me to do, and it got me to f(15), because it seemed somewhat obvious that it would.
15f(16)+f(33)=225
f(2)+f(5)=1
2f(3)+f(7)=4
3f(4)+f(9)=9
4f(5)+f(11)=16
4f(2)+4f(5)=4
f(11)-4f(2)=12
5f(6)+f(13)=25
6f(7)+f(15)=36
12f(3)+6f(7)=24
f(15)-12f(3)=12
7f(8)+f(17)=49
8f(9)+f(19)=64
24f(4)+8f(9)=72
f(19)-24f(4)=-8
f(3)=0, f(7)=4, f(15)=12
f(x)=x-3
I solved f(15) in the second method, but I wondered if we could determine a closed form solution for f(x). The way you derived f(x) = x - 3 is very clever! Great video!
Thanks
Unless you assume that f is continous. There are infinitely many solutions
@FranciszekKlyk Good point. Assuming f(1) is finite, are x = -1 and x = (2^k - 1) for integer k > 1 the only values of x for which we can unambiguously determine f(x) without further boundary conditions?
12
problem
x f(x+1) + f(2x+3) = x²
Assume f(x) is a polynomial.
Solve by a method of unknown coefficients.
Let
f(x) = a x² + b x + c
Replace in the equation.
x [a (x + 1)² + b(x + 1)+ c] +
a (2x+3)² + b (2x +3) + c = x²
Expand.
a x³ + (6a + b - 1) x² +
(7a + 3b + c) x + 9a + 3b + c = 0
Solve coefficient system of equations.
a = 0
b - 1 = 0
3 b + c = 0
b = 1
c = -3
f(x) = x - 3
f(15) = 12
answer
f(15) = 12
I think you should show why you choose certain values of x rather than just saying if x=2 then ...
but why 2.
my method ...
xf(x+1) +f(2x+3)= x^2
finding f(13)
we note the recursive nature and hope that using reducing values of the argument of f will give us a solution.
f(2x+3)= x^2-xf(x+1)
choose x to.get f(15) on.lhs
x=6
f(15)=36-6f(7)
now choose x to get f7) on the lhs
x=2
f(7)=4-2f(3)
now choose x to get f(3) on lhs
x=0
f(3) =0
substitue back
f(7)=4 -0
f(13)=36-6f(7)
f(13)=36-24 =12
in thia way we dont just guess values of z
we calculate them.
if you didn't spot that another approach would be to try x=0,1,2,3,... and see what you get
another approach is to change the dummy variable(x)
eg choose a=2x+3 or a=x+1. doesnt gain anything here.
but cant assume linear
x*f(x+1) + f(2x+3) = x²
==>
f(2x+3) = x² - x*f(x+1)
f(15) = f(2*6+3) =
= 6² - 6*f(6+1)
= 6² - 6*f(7)
= 6² - 6*f(2*2+3)
= 6² - 6*[ 2² - 2*f(2+1) ]
= 6² - 6*[ 2² - 2*f(3) ]
= 6² - 6*[ 2² - 2*f(2*0 + 3) ]
= 6² - 6*[ 2² - 2*[ 0² - 0*f(0+1) ] ]
= 6² - 6*[ 2² - 2*[ 0² - 0*f(1) ] ]
= 6² - 6*[ 2² - 2*[ 0 ] ]
= 6² - 6*[ 2² - 0 ]
= 6² - 6*[ 4 ]
= 6² - 24
= 36 - 24
= 12