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If f(0) =1, f(x) =x+f(x-1) for x>0 calculate f(100)?
- Thread starter edonag
- Start date
Okay, so you've got a sequence. It's recursively defined, which means that each term in the sequence depends on the previous one. You're given the initial term, f(0)=1. Do you remember what this function notation means? So, using the definition given to you of f(x), can you find f(1)? Can you find f(2)? Proceed onward until you find f(100).
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If f(0) =1, f(x) =x+f(x-1) for x>0 calculate f(100)?
help me
f(1) = 1 + f(0) = 2 = (1 + 1)
f(2) = 2 + f(1) = 2 + 2 = (1+2)+1
f(3) = 3 + f(2) = 3 + 4 = (1+2+3)+1
f(4) = 4 + f(3) = 7 + 4 = (1+2+3+4)+1
What are your thoughts?
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help me
Sometimes the way to tackle a problem like this is to see if you can reduce it to something you know. So lets start with
f(x) = x + f(x-1)
but f(x-1) = x-1 + f(x-2) so
f(x) = x + x-1 + f(x-2)
but f(x-2) = x-2 + f(x-3) so
f(x) = x + x-1 + x-2 + f(x-3)
Note the the 'number' in the function (the x-3 juts above) is always 1 less that the 'number' just before it. So, going until "the 'number' just before it" is, say, 2 and we have
f(x) = x + x-1 + x-2 + ... + 2 + f(1)
f(x) = x + x-1 + x-2 + ... + 2 + 1 + f(0)
f(x) = x + x-1 + x-2 + ... + 2 + 1 + 1
We have a closed form for the sum of the first x integers. Lets call that S(x) and we have
f(x) = S(x) + 1
thank you this is how i solved it!
f(0)=1
f(1)=1+f(0)
f(2)=2+f(1)
f(3)=3+f(2)
...
#f(100)=1+2+3+...+100+f(0)
all this gives us
S100=(100*101)/2 thanks to the formula Sn=(n*(n+1))/2
S100=5050
we replace 1+2+3+...+100 with 5050 at #f(100)=1+2+3+...+100+f(0)and now we have:
f(100)=5050+f(0)
f(100)=5050+1=5051
thank you for all your answers you helped me a lot!
f(0)=1
f(1)=1+f(0)
f(2)=2+f(1)
f(3)=3+f(2)
...
#f(100)=1+2+3+...+100+f(0)
all this gives us
S100=(100*101)/2 thanks to the formula Sn=(n*(n+1))/2
S100=5050
we replace 1+2+3+...+100 with 5050 at #f(100)=1+2+3+...+100+f(0)and now we have:
f(100)=5050+f(0)
f(100)=5050+1=5051
thank you for all your answers you helped me a lot!