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 nthn^{\text{th}} term of such a sequence is given by an=a+(n1)da_n=a+(n-1)d, where aa and dd 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
4 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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 - 4 years, 1 month ago

Log in to reply

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 - 4 years, 1 month ago

Log in to reply

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 - 4 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...