If the set is a complex linear space?
$begingroup$
The set $P$ consists of complex trigonometric polynomials $$f_k(t)=sum_{j=0}^k a_j e^{iτ_jt},$$ where $$ kin mathbb N, a_1,...,a_k in mathbb C, τ_1,...,τ_k in mathbb R.$$
Show that $P$ is a complex linear space.
I have to show that $$f,g in P, α, β in mathbb C Rightarrow (αf+βg) in P.$$
My solution:
$$(αf+βg)_k(t) = sum_{j=0}^k (αa_j e^{iτ_jt} + βa_j e^{iτ_jt} ) = sum_{j=0}^k αa_j e^{iτ_jt} + sum_{j=0}^k βa_j e^{iτ_jt} \ = αsum_{j=0}^k a_j e^{iτ_jt} + βsum_{j=0}^k a_j e^{iτ_jt} = αf_k(t) + βg_k(t). $$
And it is true.
Question:
Is my solution correct?
functional-analysis polynomials complex-numbers
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
asked Dec 11 '18 at 16:24
LoreenLoreen
83
$endgroup$
add a comment |
$begingroup$
The set $P$ consists of complex trigonometric polynomials $$f_k(t)=sum_{j=0}^k a_j e^{iτ_jt},$$ where $$ kin mathbb N, a_1,...,a_k in mathbb C, τ_1,...,τ_k in mathbb R.$$
Show that $P$ is a complex linear space.
I have to show that $$f,g in P, α, β in mathbb C Rightarrow (αf+βg) in P.$$
My solution:
$$(αf+βg)_k(t) = sum_{j=0}^k (αa_j e^{iτ_jt} + βa_j e^{iτ_jt} ) = sum_{j=0}^k αa_j e^{iτ_jt} + sum_{j=0}^k βa_j e^{iτ_jt} \ = αsum_{j=0}^k a_j e^{iτ_jt} + βsum_{j=0}^k a_j e^{iτ_jt} = αf_k(t) + βg_k(t). $$
And it is true.
Question:
Is my solution correct?
functional-analysis polynomials complex-numbers
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
asked Dec 11 '18 at 16:24
LoreenLoreen
83
$endgroup$
$begingroup$
Yes, it looks just fine...with a somewhat funny notation, but fine.
$endgroup$
– DonAntonio
Dec 11 '18 at 16:26
$begingroup$
Well, I think the choice of $k$ and of $tau_0,tau_1,dots,tau_k$ varies. That is, not all elements of $P$ use the same ones.
$endgroup$
– GEdgar
Dec 11 '18 at 18:38
add a comment |
$begingroup$
The set $P$ consists of complex trigonometric polynomials $$f_k(t)=sum_{j=0}^k a_j e^{iτ_jt},$$ where $$ kin mathbb N, a_1,...,a_k in mathbb C, τ_1,...,τ_k in mathbb R.$$
Show that $P$ is a complex linear space.
I have to show that $$f,g in P, α, β in mathbb C Rightarrow (αf+βg) in P.$$
My solution:
$$(αf+βg)_k(t) = sum_{j=0}^k (αa_j e^{iτ_jt} + βa_j e^{iτ_jt} ) = sum_{j=0}^k αa_j e^{iτ_jt} + sum_{j=0}^k βa_j e^{iτ_jt} \ = αsum_{j=0}^k a_j e^{iτ_jt} + βsum_{j=0}^k a_j e^{iτ_jt} = αf_k(t) + βg_k(t). $$
And it is true.
Question:
Is my solution correct?
functional-analysis polynomials complex-numbers
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
asked Dec 11 '18 at 16:24
LoreenLoreen
83
$endgroup$
The set $P$ consists of complex trigonometric polynomials $$f_k(t)=sum_{j=0}^k a_j e^{iτ_jt},$$ where $$ kin mathbb N, a_1,...,a_k in mathbb C, τ_1,...,τ_k in mathbb R.$$
Show that $P$ is a complex linear space.
I have to show that $$f,g in P, α, β in mathbb C Rightarrow (αf+βg) in P.$$
My solution:
$$(αf+βg)_k(t) = sum_{j=0}^k (αa_j e^{iτ_jt} + βa_j e^{iτ_jt} ) = sum_{j=0}^k αa_j e^{iτ_jt} + sum_{j=0}^k βa_j e^{iτ_jt} \ = αsum_{j=0}^k a_j e^{iτ_jt} + βsum_{j=0}^k a_j e^{iτ_jt} = αf_k(t) + βg_k(t). $$
And it is true.
Question:
Is my solution correct?
functional-analysis polynomials complex-numbers
functional-analysis polynomials complex-numbers
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
asked Dec 11 '18 at 16:24
LoreenLoreen
83
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
asked Dec 11 '18 at 16:24
LoreenLoreen
83
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
edited Dec 11 '18 at 17:53
Paul Frost
11.8k3935
11.8k3935
asked Dec 11 '18 at 16:24
LoreenLoreen
83
asked Dec 11 '18 at 16:24
LoreenLoreen
83
asked Dec 11 '18 at 16:24
LoreenLoreen
83
83
$begingroup$
Yes, it looks just fine...with a somewhat funny notation, but fine.
$endgroup$
– DonAntonio
Dec 11 '18 at 16:26
$begingroup$
Well, I think the choice of $k$ and of $tau_0,tau_1,dots,tau_k$ varies. That is, not all elements of $P$ use the same ones.
$endgroup$
– GEdgar
Dec 11 '18 at 18:38
add a comment |
$begingroup$
Yes, it looks just fine...with a somewhat funny notation, but fine.
$endgroup$
– DonAntonio
Dec 11 '18 at 16:26
$begingroup$
Well, I think the choice of $k$ and of $tau_0,tau_1,dots,tau_k$ varies. That is, not all elements of $P$ use the same ones.
$endgroup$
– GEdgar
Dec 11 '18 at 18:38
$begingroup$
Yes, it looks just fine...with a somewhat funny notation, but fine.
$endgroup$
– DonAntonio
Dec 11 '18 at 16:26
$begingroup$
Yes, it looks just fine...with a somewhat funny notation, but fine.
$endgroup$
– DonAntonio
Dec 11 '18 at 16:26
$begingroup$
Well, I think the choice of $k$ and of $tau_0,tau_1,dots,tau_k$ varies. That is, not all elements of $P$ use the same ones.
$endgroup$
– GEdgar
Dec 11 '18 at 18:38
$begingroup$
Well, I think the choice of $k$ and of $tau_0,tau_1,dots,tau_k$ varies. That is, not all elements of $P$ use the same ones.
$endgroup$
– GEdgar
Dec 11 '18 at 18:38
add a comment |
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$begingroup$
Yes, it looks just fine...with a somewhat funny notation, but fine.
$endgroup$
– DonAntonio
Dec 11 '18 at 16:26
$begingroup$
Well, I think the choice of $k$ and of $tau_0,tau_1,dots,tau_k$ varies. That is, not all elements of $P$ use the same ones.
$endgroup$
– GEdgar
Dec 11 '18 at 18:38