Please show proper notation, justification and step by step work. n See attachment for problem

Given that the zeros for (sinx)/x are the values x=0, x= +-pie, x=+-2pie, x=+-3pie, . (x=0 must be excluded, why?)

This implies that F(x) can be factored as follows

F(x) = (1-(x/pi)) (1-(x/-pi)) (1-(x/2pi)) (1-(x/3pi)) (1-(x/-3pi)) .

= [(1-(x/pi))(1+(x/pi))] [(1-(x/2pi))(1+(x/2pi))] [(1-(x/3pi))(1+(x/3pi)) .

= (1-((x^2)/(pi^2))) (1-((x^2)/(4pi^2))) (1-((x^2)/(9pi^2))) (1-((x^2)/(16pi^2))) (1- ((x^2)/(25pi^2)))

Multiply out this infinite product (or at least the first 6 factors) and collect like terms in x's to form a polynomial.

i.e F(x)= 1- (???????)x^2 + (????????)x^4 -(???????)x^6 + ..

My notation is very bad but it is the best I could do with my version of

windows. Please show step by step work and a nice tidy answer.

Given that the zeros for (sinx)/x are the values x=0, x= +-pie,

x=+-2pie, x=+-3pie,â¦. (x=0 must be excluded, why?)

This implies that F(x) can be factored as follows

F(x) = (1-(x/pi)) (1-(x/-pi)) (1-(x/2pi)) (1-(x/3pi))

(1-(x/-3pi))â¦.

= [(1-(x/pi))(1+(x/pi))] [(1-(x/2pi))(1+(x/2pi))]

[(1-(x/3pi))(1+(x/3pi))â¦.

= (1-((x^2)/(pi^2))) (1-((x^2)/(4pi^2))) (1-((x^2)/(9pi^2)))

(1-((x^2)/(16pi^2))) (1- ((x^2)/(25pi^2)))â¦

Multiply out this infinite product (or at least the first 6 factors) and

collect like terms in xâs to form a polynomial.

i.e F(x)= 1- (???????)x^2 + (????????)x^4 -(???????)x^6 +â¦â¦..

Given that the zeros for (sinx)/x are the values x=0, x= +-pie, x=+-2pie, x=+-3pie, . (x=0 must be excluded, why?)

This implies that F(x) can be factored as follows

F(x) = (1-(x/pi)) (1-(x/-pi)) (1-(x/2pi)) (1-(x/3pi)) (1-(x/-3pi)) .

= [(1-(x/pi))(1+(x/pi))] [(1-(x/2pi))(1+(x/2pi))] [(1-(x/3pi))(1+(x/3pi)) .

= (1-((x^2)/(pi^2))) (1-((x^2)/(4pi^2))) (1-((x^2)/(9pi^2))) (1-((x^2)/(16pi^2))) (1- ((x^2)/(25pi^2)))

Multiply out this infinite product (or at least the first 6 factors) and collect like terms in x's to form a polynomial.

i.e F(x)= 1- (???????)x^2 + (????????)x^4 -(???????)x^6 + ..

My notation is very bad but it is the best I could do with my version of

windows. Please show step by step work and a nice tidy answer.

Given that the zeros for (sinx)/x are the values x=0, x= +-pie,

x=+-2pie, x=+-3pie,â¦. (x=0 must be excluded, why?)

This implies that F(x) can be factored as follows

F(x) = (1-(x/pi)) (1-(x/-pi)) (1-(x/2pi)) (1-(x/3pi))

(1-(x/-3pi))â¦.

= [(1-(x/pi))(1+(x/pi))] [(1-(x/2pi))(1+(x/2pi))]

[(1-(x/3pi))(1+(x/3pi))â¦.

= (1-((x^2)/(pi^2))) (1-((x^2)/(4pi^2))) (1-((x^2)/(9pi^2)))

(1-((x^2)/(16pi^2))) (1- ((x^2)/(25pi^2)))â¦

Multiply out this infinite product (or at least the first 6 factors) and

collect like terms in xâs to form a polynomial.

i.e F(x)= 1- (???????)x^2 + (????????)x^4 -(???????)x^6 +â¦â¦..

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