Taylor series formula at a point

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What is the taylor series formula for f(x)= cosx at x= pi/4.

Apr 27th, 2015

Taylor series 

cos(x) = f (a)(x-a)^0/0! + f ' (a)(x-a)^1/1! + f '' (a)(x-a)^2/2! + f ''' (a)(x-a)^3/3! + f '''' (a)(x-a)^4/4! + .............

 where  a = π/4 

f (a) = cosx 
f ' (a) = -sinx 
f '' (a) = -cosx 
f ''' (a) = sinx 
f '''' (a) = cosx 

f (π/4) = 1/√2 
f ' (π/4) = -1/√2 
f '' (π/4) = -1/√2 
f ''' (π/4) = 1/√2 
f '''' (π/4) = 1/√2 

now put the values in  Taylor series .

cos(x) = 1/√2 - 1/√2(x-π/4) - 1/√2(x-π/4)^2/2 + 1/√2(x-π/4)^3/6 + 1/√2(x-π/4)^4/24 + ....


Apr 27th, 2015

any doubt let me know 

Apr 27th, 2015

Thanks a lot.

Apr 27th, 2015

welcome here ..

if you have more question you may  ask directly to me

Apr 27th, 2015

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