anonymous
  • anonymous
The function satisfies the identity f(x) + f(y) = f(x+y) (1) for all x and y. Show that 2f(x) = f(2x) and deduce that f '' (0) = 0. By considering the Maclaurin series for f(x), find the most general function that satisfies (1). [Do not consider issues of existence or convergence of Maaclaurin series in this question.]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@TuringTest help me! :P
anonymous
  • anonymous
@experimentX
JamesJ
  • JamesJ
First, it is clear to you that f(2x) = 2f(x), yes?

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JamesJ
  • JamesJ
talk to me, I'll help you, but you need to be here.
anonymous
  • anonymous
yes it is, sorry I wasn't there. But everything after that I am not ok with. How do I deduce that f''(0) = 0 ?
JamesJ
  • JamesJ
Differentiate the equation f(2x) = 2f(x) twice
anonymous
  • anonymous
2f'(2x) = 2f'(x) 4f''(2x) = 2f''(x) ?
JamesJ
  • JamesJ
Right. Now evaluate that at zero and what do you get?
anonymous
  • anonymous
uhh 4f''(0) = 2f''(0) or you mean when f''(x) = 0 ? Sorry I'm probably being stupid
JamesJ
  • JamesJ
Yes. Notice that all of the higher derivatives at x = 0 are also zero.
anonymous
  • anonymous
Oh because f''(0)=0 is the only solution to 4f''(0) = 2f''(0) ?
anonymous
  • anonymous
so I can see anyway now that there will only be 2 terms in the maclaurin series, the rest will be zero
JamesJ
  • JamesJ
Yes, that shows that f''(0) = 0. Now keep on differentiating the original expression and you can show that \[ f^{(n)}(0) = 0 \] i.e., the nth derivative of f is also zero when x = 0.
anonymous
  • anonymous
ok
anonymous
  • anonymous
so is it just ax + b?
JamesJ
  • JamesJ
Those are the only possibilities at this stage. Now put that into the original equation and see if you can say anything else.
anonymous
  • anonymous
2ax + b = 2(ax + b) so it's not possible unless b = 0 ?
JamesJ
  • JamesJ
Right. Alternatively, ask yourself this: can we say anything about f(0)? Yes, it must be ...
anonymous
  • anonymous
Wow, you're great :)
JamesJ
  • JamesJ
My mother says so too. But thanks.
anonymous
  • anonymous
Hah :) thank you

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