If the set is a complex linear space?












1












$begingroup$


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?










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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
















1












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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edited Dec 11 '18 at 17:53









Paul Frost

11.1k3934




11.1k3934










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


















  • $begingroup$
    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










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