# Problem with roots of a function h(x)

1. Aug 6, 2005

### eljose

Let be the function h(x)=f(x)+g(x) we want to obtain the values of x so h(x)=0 and we have that f(x) and g(x) have the same roots (if a is so f(a)=0 then g(a)=0 too) so we have two types of roots:

1.-values of x that satisfy f(x)=-g(x)
2.values x that make f(x)=g(x)=0

then we make in the first equation f(x)/g(x)=-1 this should be valid for any root of h(x) the problem would be if the roots of f(x) and g(x) would also satisfy the formula f(a)/g(a)=-1 with a a root of f(x) and g(x) ,we have that the values that satisfy the equation h(x)=0->that f(x)/g(x)=-1 so as a is a root of f and g would also satisfy that h(a)=0=0+0 and so should satisfy that f(x)/g(x)=-1 but how to prove this?...thanks.

2. Aug 6, 2005

### LeonhardEuler

If a is a root of both f and g, f(a)/g(a) is not equal to -1. It is 0/0, which is meaningless. It may or may not be the case that $$\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=-1$$

3. Aug 6, 2005

### Hurkyl

Staff Emeritus
I'm having a very difficult time trying to parse what you're saying.

It almost sounds as if you're suggesting that, if a is a root of h, then it is always true that f(a) / g(a) = -1, but that's clearly false, by the very observation you made, that it might be the case that f(a) = g(a) = 0.

4. Aug 7, 2005

### eljose

but we may have that the limit Lim x->a [f(x)/g(x)]=1 with []=modulus of the function.

another question,let,s suppose we have the function h(x)=F(a+ix)e^{-iwx}+F(a-ix)e^{iwx} so the function is real,now we want to obtain the roots of h(x)=0 F(x) is real so F(a+ix) and F(a-ix) are one the conjugate of other let,s suppose c is a root of both F(a+ix) and F(a-ix) so F(a+ic)=F(a-ic)=0 then would we have that the limit when x-> [F(a+ix)/F(a-ix)]=1..

my reasioning is that expanding F(a+ix) and F(a-ix) near the root c and supposing F(a+ix) and F(a-ix) don,t have an extremum at x=c we have that:

F(a+ix)=F(a+ic)i(x-c) and F(a-ix)=F´(a-ic)(x-c)(-i) so dividing one by other we would have that Limx->c F(a+ix)/F(a-ix)=(-1)F(a+ic)/F´(a-ic) but we also have that ^F´(a+ic) and F`(a-ic) are conjugate so their quotient will be e^{id} for some real d but taking the modulus of [F´(a+ic)/F´(a-ic)(-1)]=1 as we wanted to prove....

5. Aug 7, 2005

### Hurkyl

Staff Emeritus
Please use correct punctuation. It greatly helps readability. :tongue2: e.g. periods at ends of sentences, and writing apostrophes as apostrophes. (It's "let's", not "let,s")

If h is given by $h(x)=F(a+ix)e^{-iwx}+F(a-ix)e^{iwx}$, that is not sufficient to conclude that h is real.

If b is a root of h, that does not imply that F(a + ib) = 0.

Also, when z is nonzero, we have the identity $|z / \bar{z}| = 1$.

6. Aug 7, 2005

### eljose

thanks for your reply, but if F(x) is real then F(a+ix) and F(a-ix) (if we set w=0) then F(a+ix) would be the complex conjugate of F(a-ix) and viceversa (expanding both functions in x would give the answer)....

if b is a root of h this doesn,t not imply that F(a+ib)=0 but you will agree that the set of roots that satisfy F(a+ib)=0 is included in the set of roots of h(x) (if b is a root of F(a+ix) then is also a root of h(x)=0)

7. Aug 7, 2005

8. Aug 8, 2005

### eljose

but you will have that f(a+it)=R(t)e^{iv(t)} and f(a-it)=R(t)e^{-iv(t)} with t real and v(t) takes real values...

Edit:well i think that the f(a+it)+f(a-it) will only be real if $$f^{*}(z)=f(z^*)$$ with *=complex conjugate....

tehn if f(a+it)+f(a-it) is real then will be true that f(a+it)?R(t)exp{iv(t)} and
f(a-it)=R(t)exp{-iv(t)} so the sum will be real...

then if f(a+it)+f(a-it)=z(t) is real then its derivatives D^{n}Z(t) will also be real so we will be able to express D^{n}Z(t)=W_n(t)exp{iU_n(t)} for the sets of function U_n(t) and W_n(t).

Last edited: Aug 8, 2005
9. Aug 8, 2005

### matt grime

you' ve not answered any of the questions asked of you. Let us show you by example:

Let F(z)=i if im(z)=/=0 and 1 if z is purely real.

then for any real x, F(x) is real but F(a+ix) and F(a-ix) do not satisfy your claim of complex conjugacy nor of h(x) being a real valued function, indeed h(2pi)=2i

why don't you take care and write things out properly explaining all the assumptions you are making? What, for instance is R(t), and v(t) now? not to mention all the other undefined terms you've now added to your edited post

Last edited: Aug 8, 2005
10. Aug 8, 2005

### matt grime

but you didn't specify this requirement originally, did you? and it isn't even true ofr f analytic. it would be true if, say there were a taylor series convergent everywhere and with purely real coefficients.

11. Aug 8, 2005

### eljose

and let be the f with f real that have f(a+ib)=exp(ik) and f(c+id)=exp(-ik) then should we have necessaary that a=c and b=-d (k is the same for both cases)....

12. Aug 8, 2005

### matt grime

what on earth are you on about now? where did the k come from? f is now a function from C to R?... madness

13. Aug 8, 2005

### eljose

k is a real constant we have that f(a+ib) and f(c+id) have the same modulus....then my reasoning is that if we express f(a+ib) and f(c+id) in terms of their real and complex part we would have that f(a+ib)=U(a,b)+iV(a,b) and f(c+id)=U(c,d)+iV(c,d) then as their modulus is 1 then we have that U(a,b)=U(c,d) and V(a,b)=V(c,d)

we have also that their complex part is the same but a sign so we would have that...
Artan[V(a,b)/U(a,b)]=-Artan[V(c,d)/U(c,d)] from this i think we could deduce that a=c

14. Aug 8, 2005

### matt grime

if this reflects your standard of presentation in these things you write then no wonder they are rejected. even now no one is any the wiser as to what you mean. none of what you have written can with any reasonable certainty be said to ve true, it is almost surely false. think it's time to stop wasting time in working out what you're trying to say. try again when you can make it readable.

15. Aug 8, 2005

### eljose

i have another question let be $$[f(a+ib)]^*$$ the complex conjugate of the function $$f(a+ib)$$ then my question is if we denote f^{-1} the inverse function of f i would like to know if the equality holds...

$$f^{-1}*(f^*(a+ib))=a+ib$$ that is the conjugate of the inverse function applied to the conjugate function is the x function.....

16. Aug 8, 2005

### matt grime

why are youi asking? you don't bother to listen to any replies.