Another way of looking at A.P. (Derivatives)

An Arithmetic Progression is defined as a sequence such that the difference between two consecutive terms is constant. Hence the \(n^{\text{th}}\) term of such a sequence is given by \(a_n=a+(n-1)d\), where \(a\) and \(d\) denote the first term and the common difference respectively. So I was looking at an A.P. with this common difference as the rate of change of a term of an A.P..

If the nthn^{\text{th}} term of an A.P. is given by an=an+ba_n=an+b, then we have

a1=a+b and a2=2a+bk=a2a1=ak=aa_1=a+b~\text{and}~a_2=2a+b \\ \Rightarrow k=a_2-a_1=a \\ \boxed{k=a}

where kk denotes the common difference. This is in sync with

k=dandn=ddn(an+b)=ak=ak=\dfrac{da_n}{dn}=\dfrac{d}{dn}(an+b)=a \\ \boxed{k=a}

But it doesn't go hand in hand here if we look at the sums of 1,2,3,1,2,3,\dots terms of an A.P. from the beginning changing at the rate of the general term ana_n. If Sn=an2+bnS_n=an^2+bn denotes the sum of first nn terms of an A.P. (it will be of this form as an existent formula for the sum of first nn terms of an A.P. which can be rearranged into this form is n2(2a+(n1)d)\dfrac{n}{2}(2a+(n-1)d), where aa and dd are the first term and the common difference of an A.P. respectively), then we have

a1=S1=a+b and a1+a2=S2=4a+2ba2=S2S1=3a+bk=a2a1=2aan=(a+b)+(n1)(2a)=ba+2anan=ba+2ana_1=S_1=a+b~\text{and}~a_1+a_2=S_2=4a+2b \\ \Rightarrow a_2 = S_2-S_1=3a+b \\ \Rightarrow k=a_2-a_1=2a \\ \Rightarrow a_n=(a+b)+(n-1)(2a)=b-a+2an \\ \boxed{a_n=b-a+2an}

And if we go with derivatives, we have

an=dSndn=ddn(an2+bn)=2an+ban=2an+ba_n=\dfrac{dS_n}{dn}=\dfrac{d}{dn}(an^2+bn)=2an+b \\ \boxed{a_n=2an+b}

Can someone explain to me the disappearance of the a-a? One interesting observation is that this does hold true.

k=dandn=ddn(an)=ddn(dSndn)=d2Sndn2k=d2dn2(an2+bn)=2ak=2ak=\dfrac{da_n}{dn}=\dfrac{d}{dn}(a_n)=\dfrac{d}{dn}\left(\dfrac{dS_n}{dn}\right)=\dfrac{d^2S_n}{dn^2} \\ k = \dfrac{d^2}{dn^2}(an^2+bn)=2a \\ \boxed{k=2a}

Note by Omkar Kulkarni
5 years, 10 months ago

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1 vote

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You can't differentiate it that way. It's the same reason why you can't differntiate x2=x+x+x++xx^2 = x + x + x +\ldots + x to get 2x=1+1+1++12x = 1 + 1 + 1 + \ldots + 1 .

Pi Han Goh - 5 years, 10 months ago

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Sorry, lot of questions. I'm very much new to differentiation. 1. Why can't we differentiate in your example? 2. Is the fact that the other two things hold true a coincidence?

Omkar Kulkarni - 5 years, 10 months ago

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What I have is a discrete case. The equation is only true for discrete number of xx. You need to make sure it's continuous first before differentiating. No, it's not a coincidence, it's just the way it was set up to mislead people.

Pi Han Goh - 5 years, 10 months ago

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