Proving the Minimum Radius of Convergence for a Sum of Taylor Series

[jaykobe76]

Homework Statement

how to prove that radius of convergence of a sum of two series is greater or equal to the minimum of their individual radii

i don't know how to begin, can someone give me some ideas?

 

[kai_sikorski]

Hi Jay,

Welcome to the Forums!

I wouldn't try to do this by working with the coefficients of the series directly. Instead fix some x within the radius of convergence of both sequences. Define f_n(x) to be the quantity that x gets mapped to by the first n terms of the first power series, and g_n(x) the same thing except for the second series. What do you know about the sequences

f₁(x), f₂(x), f₃(x), ... and
g₁(x), g₂(x), g₃(x), ... ?

Using that what can you say about the sequence (g + f)_n(x)?

 

[jaykobe76]

Hi Jay,

Welcome to the Forums!

I wouldn't try to do this by working with the coefficients of the series directly. Instead fix some x within the radius of convergence of both sequences. Define f_n(x) to be the quantity that x gets mapped to by the first n terms of the first power series, and g_n(x) the same thing except for the second series. What do you know about the sequences

f₁(x), f₂(x), f₃(x), ... and
g₁(x), g₂(x), g₃(x), ... ?

Using that what can you say about the sequence (g + f)_n(x)?

now , i got some ideas. can i say that cause fn(X) gn(x) are both converges to somthing ,assume T ,S .then by definition of convergent we have |fn(x)-T|<esillope/2 for |x|<R1,and |gn(x）-S|<esillope/2 for |x|<R2 (R1 R2 are the radius of f(X) and g(X)) then choose the min{R1,R2} then we can get |fn(X)+gn(x)-(T+S)|<esillope which is convergent for |x|<min{R1,R2},then the new radius K=min{R1,R2} but how to conclude that it is greater than min{R1,R2}??and is the proof above correct?

 

[kai_sikorski]

Above proof is correct.

I don't see how the statement that the radius of convergence of the sum is strictly greater than the min(R₁,R₂) is even true. I mean take the taylor series for g(x)=0 as one of the series. Clearly the radius of convergence of the sum must be exactly equal to the radius of convergence of the other series. So greater than or equal to is best you can hope for.

 

[kai_sikorski]

Oh if you're asking why it might be greater, then it's simple because your proof excluded the possibility that

R_{1 + 2} < min(R₁, R₂)

So it must be true that

R_{1 + 2} ≥ min(R₁, R₂)

That means that it might be greater maybe, or it might be equal. And in fact you could find examples for both.

 

[jaykobe76]

Oh if you're asking why it might be greater, then it's simple because your proof excluded the possibility that

R_{1 + 2} < min(R₁, R₂)

So it must be true that

R_{1 + 2} ≥ min(R₁, R₂)

That means that it might be greater maybe, or it might be equal. And in fact you could find examples for both.

sorry, i still cannot understand. from my proof i concluded that |x|<min{R1,R2}then the new radius =min{R1,R2}=K. then the sum of series convergent.but if K can be greater than min{R1,R2}then |x|<min{R1,R2}=<K ,but how can i get this from the proof?

 

[jaykobe76]

Oh if you're asking why it might be greater, then it's simple because your proof excluded the possibility that

R_{1 + 2} < min(R₁, R₂)

So it must be true that

R_{1 + 2} ≥ min(R₁, R₂)

That means that it might be greater maybe, or it might be equal. And in fact you could find examples for both.

sorry, i think i get it . because from my proof that |x|<min{R1,R2} then it means that the new radius K cannot smaller than {R1,R2}so K>=min{R1,R2} is this the idea?

 

[kai_sikorski]

sorry, i think i get it . because from my proof that |x|<min{R1,R2} then it means that the new radius K cannot smaller than {R1,R2}so K>=min{R1,R2} is this the idea?

Yes exactly.

 

[kai_sikorski]

then by definition of convergent we have |fn(x)-T|<esillope/2 for |x|<R1

I guess here technically you should say that there exists an N, such that this is true for all fn(x) with n>N.

 

[jaykobe76]

I guess here technically you should say that there exists an N, such that this is true for all fn(x) with n>N.

ok thank you so much

 

[jaykobe76]

I guess here technically you should say that there exists an N, such that this is true for all fn(x) with n>N.

sorry, can use give me one example that the T>min{R1,R2}(inequality is strict) cause i cannot find one case that satisfies.

 

[kai_sikorski]

just take a taylor series for some function f(x) that has finite radius of convergence. make g(x) the taylor series for -f(x).