Continuous functions Watch

Feynboy
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I got a) quickly, f(0)=0.

For b) it suggests "first establishing why
f(n) = n f(1) for n ∈ N, then seeing why f( n/ m ) = n/m f(1) for n, m ∈ N and finally dealing with the negative rational numbers."

I showed f(n)=nf(1)

How can I do the rest? I did f(n) by representing n as summation of 1 and getting n f(1). I can't do this for fractions.

I tried breaking n/m=a+b/c where b/c<1. But can't find a way
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DFranklin
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(Original post by Feynboy)
Attachment 803122
I got a) quickly, f(0)=0.

For b) it suggests "first establishing why
f(n) = n f(1) for n ∈ N, then seeing why f( n/ m ) = n/m f(1) for n, m ∈ N and finally dealing with the negative rational numbers."

I showed f(n)=nf(1)

How can I do the rest? I did f(n) by representing n as summation of 1 and getting n f(1). I can't do this for fractions.

I tried breaking n/m=a+b/c where b/c<1. But can't find a way
For any x, you can show f(nx) = n f(x) (just as you did when x = 1).
Then set x = 1/n.

Edit: Sorry - thought you were just stuck on finding f(1/n). You can also set x = m/n to find f(m/n) similarly.
Last edited by DFranklin; 1 week ago
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Feynboy
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Great, thank you
(Original post by DFranklin)
For any x, you can show f(nx) = n f(x) (just as you did when x = 1).
Then set x = 1/n.

Edit: Sorry - thought you were just stuck on finding f(1/n). You can also set x = m/n to find f(m/n) similarly.
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Feynboy
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(Original post by DFranklin)
For any x, you can show f(nx) = n f(x) (just as you did when x = 1).
Then set x = 1/n.

Edit: Sorry - thought you were just stuck on finding f(1/n). You can also set x = m/n to find f(m/n) similarly.
I showed this for any x, however then c) asks to prove f(x)=xf(1) for any x. Does proving f(nx)=nf(x) also assume the function is continuous?
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DFranklin
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(Original post by Feynboy)
I showed this for any x, however then c) asks to prove f(x)=xf(1) for any x. Does proving f(nx)=nf(x) also assume the function is continuous?
No. The key point here is that n is an integer, not an arbitrary real. So for n > 0, it's just the observation that f(nx) = f(\underbrace{x+x+...+x}_{\text {n terms}}) = \underbrace{f(x)+f(x)+...+f(x)}_  {\text{n terms}} = nf(x). You then need to deal with the n =0 and n < 0 cases (not hard).

[i.e. This should be exactly the same argument you used for f(n) = n f(1), only with x replacing 1].
Last edited by DFranklin; 1 week ago
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Feynboy
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(Original post by DFranklin)
No. The key point here is that n is an integer, not an arbitrary real. So for n > 0, it's just the observation that f(nx) = f(\underbrace{x+x+...+x}_{\text {n terms}}) = \underbrace{f(x)+f(x)+...+f(x)}_  {\text{n terms}} = nf(x). You then need to deal with the n =0 and n < 0 cases (not hard).

[i.e. This should be exactly the same argument you used for f(n) = n f(1), only with x replacing 1].
I see, seems like the guide to the question made it much more complicated than this wayName:  Screenshot 2019-03-14 at 17.21.42.png
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Feynboy
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Any idea with (c)? I know f(q_n))_n=1 ^infinity ---> f(x) since f is continuous
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DFranklin
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(Original post by Feynboy)
Any idea with (c)? I know f(q_n))_n=1 ^infinity ---> f(x) since f is continuous
Yes. So how might this help...? (It is very hard to give a hint here without just answering the question).
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Feynboy
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(Original post by DFranklin)
Yes. So how might this help...? (It is very hard to give a hint here without just answering the question).
so f(q_1)+...+f(q_n) --> f(x)
so (q_1+...+q_n)f(1) --> f(x)

can we now just write an = sign?
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