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Probably easier than I think f(xn)=nf(x)?

  • Thread starter geoman
  • Start date
1. show that for each positive integer n and each real number x, f(nx)=nf(x).



2. Homework Equations
f(x) is an additive function (f(x+y)=f(x)+f(x))



3. The Attempt at a Solution
Well I'm thinking that I can just use mathematical induction to show that:
1) f((1)x)=1f(x)
2)f(nx)=nf(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+....+f[tex]_{}n[/tex](x) (assume this is true)
3)f((n+1)x)=(n+1)f(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+....+f[tex]_{}n+1[/tex](x)
from here it would just be algebra. is this right?
note: the numbered functions above are supposed to be subscripted but i messed it up somehow. They are only supposed to number the functions as I add the function n times and then n+1 times
 
Last edited:
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How you have presented this problem is confusing.
The statement you presented in section 1 is not true in general.
The statement you presented as a relevant equation is more than just relevant; it is part of the problem statement, I'm pretty sure. Also, you have a typo, and this equation should be f(x + y) = f(x) + f(y).

In your attempt at a solution, yes you should assume that f(nx) = nf(x), but how in the world do you go from nf(x) to a sum of powers of f(x)?

Look at what happens with f(x + x). What is that equal to by your original assumptions? You have a way to go before you will be able to tackle the induction step, IMO, so let's get up to speed with some simple calculations first.
 

HallsofIvy

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Homework Helper
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1. show that for each positive integer n and each real number x, f(nx)=nf(x).



2. Homework Equations
f(x) is an additive function (f(x+y)=f(x)+f(x))



3. The Attempt at a Solution
Well I'm thinking that I can just use mathematical induction to show that:
Yes, induction is the way to go.

1) f((1)x)=1f(x)
2)f(nx)=nf(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+....+f[tex]_{}n[/tex](x) (assume this is true)
What??? Where in the world did that horribly complicated right end come from? You haven't even defined "f1", "f2", etc.
Just use f(nx)= nf(x).

3)f((n+1)x)=(n+1)f(x)=f[tex]_{}1[/tex](x)+f[tex]_{}2[/tex](x)+....+f[tex]_{}n+1[/tex](x)
You can't use "f((n+1)x)= (n+1)f(x)", that's what you want to prove. You can use the fact that f((n+1)x)= f(nx+x)= what?

from here it would just be algebra. is this right?
note: the numbered functions above are supposed to be subscripted but i messed it up somehow. They are only supposed to number the functions as I add the function n times and then n+1 times
 

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