Solving functions: f(x, y) = x + f(x-1, y-1). If f(1, 0) = 5, then what is f(5, 2)?

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dmatzick

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The Question: A function f(x,y) of two variables has the property that f(x,y)=x + f(x-1, y-1). If f(1,0)=5, then what is the value of f(5,2)?
I know I have to substitute when the function equals an equation, but I don't understand where to start when the f(x,y) equations x plus a point... please help!:(

Thank you.
 
dmatzick said:
The Question: A function f(x,y) of two variables has the property that f(x,y)=x + f(x-1, y-1). If f(1,0)=5, then what is the value of f(5,2)?
I know I have to substitute when the function equals an equation, but I don't understand where to start when the f(x,y) equations x plus a point... please help!:(

Thank you.
Click to expand...
Can you calculate f(2,1)
 
This is where I am...

f(5,2) = 2 + f(5-1,2-1) = 2+(4,1) = (6,3)? I don't know what to do with the "2+" with the point (4,1) if that's even correct. thanks and still frustrated. :(
 
dmatzick said:
f(5,2) = 2 + f(5-1,2-1) = 2+(4,1) = (6,3)? I don't know what to do with the "2+" with the point (4,1)
Click to expand...
It's actually a good thing that you don't know how to add numbers to points because there's no way to do that (it makes no sense).

I think you made a substitution error and then misread some notation.

f(5,2) = 5 + f(5-1,2-1) = 5 + f(4,1)

Do you see the differences between what you wrote and what I wrote? :cool:
 
dmatzick said:
The Question: A function f(x,y) of two variables has the property that f(x,y)=x + f(x-1, y-1). If f(1,0)=5, then what is the value of f(5,2)?
Click to expand...

Are you sure they're asking for f(5,2) and not f(5,4)?

I know how to calculate f(5,4) but I'm having issues trying to determine f(5,2).
 
yes it's (5,2) is the value the question is asking for.

I do see my error and am glad that there is no way to add a number to a point (without it being a translation!)... so the answer is 5 + f(4,1)? The only other thing they say is "Hints: 191" but that meant nothing to me. Can you show me how you would back into it if it were (5,4)?
 
Yeah, solving for f(5, 2) is literally impossible unless there's some (unposted) information that OP knows that we don't. Per the given rule, f(5, 2) = 5 + f(4, 1). But what's f(4,1)? Well f(4, 1) = 4 + f(3, 0). So we have f(5, 2) = 5 + 4 + f(3, 0)... and on and on it goes. You can keep going indefinitely and you'll never reach a known value.
 
dmatzick said:
I do see my error and am glad that there is no way to add a number to a point (without it being a translation!)... so the answer is 5 + f(4,1)?
Click to expand...
I would say so.


The only other thing they say is "Hints: 191" but that meant nothing to me.
Click to expand...
191 does not help me, either.


Can you show me how you would back into it if it were f(5,4)?
Click to expand...
I've corrected your mistake above (in red).

Be careful with your function notation; you've made this mistake twice now.

What do you get, when you calculate f(5,4)? You will get a result in terms of f(4,3), yes?

Then, what do you get when you calculate the f(4,3) part? You'll get a result for that in terms of f(3,2), yes?

If you keep going, you'll eventually get a result in terms of f(1,0) which you're told is 5.

At that point, you can stop, and the answer will be obtained by substitution.

If you get stuck, let us know. :cool:
 
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