Summation Series to Product Series Part 1.

Here's something interesting I found out last year. I don't know if its been done before but I strongly suspect it has.

Consider a summation series of form S(n)=k=1k=nf(k)k!S(n) = \sum_{k = 1}^{k = n}\frac{f(k)}{k!} such that f(1)=1f(1) = 1.

Now let's do something unusual and create the following sequence starting with the term a2=2a_2 = 2 and then ak+1=(k+1)(ak+f(k)) a_{k+1} = (k + 1)(a_k + f(k)) for all k>2k > 2.

Finally we'll focus our attention on the following product series P(n)=k=2k=n(1+f(k)ak)P(n) = \prod_{k=2}^{k=n} ( 1 + \frac{f(k)}{a_k}) for all n>2 n > 2.

We'll show that S(n)=P(n) S(n) = P(n) for all n>2n > 2.

Proof : By induction on n. We'll start with the base case. Base Case: n=2n = 2

We have P(2)=1+f(2)a2=1+f(2)2=f(1)1+f(2)2=S(2)P(2) = 1 + \frac{f(2)}{a_2} = 1 + \frac{f(2)}{2} =\frac{f(1)}{1} + \frac{f(2)}{2} = S(2) as required.

Induction Step

This part is a quite long, so we'll break it down into 3 parts.

Part 1 : Re-arranging the Factors of the Product Series

We can write each factor from our product series as 1+f(k)ak=akak+f(k)ak=ak+f(k)ak=ak+1(k+1)ak\begin{aligned} 1 + \frac{f(k)}{a_k} &= \frac{a_k}{a_k} + \frac{f(k)}{a_k} \\&= \frac{a_k + f(k)}{a_k} \\&= \frac{a_{k+1}}{(k+1)a_k} \end{aligned} by using the definition of our sequence.

Part 2 : Re-writing our Product Series

Now we have that k=2k=n(1+f(k)ak)=a3(3)a2×a4(4)a3×...×an+1(n+1)an=an+1a2×3×4×...×n×(n+1)=an+1(n+1)! \begin{aligned} \prod_{k = 2}^{k = n} (1 + \frac{f(k)}{a_k}) &= \frac{a_{3}}{(3)a_2} \times \frac{a_{4}}{(4)a_3} \times ...\times \frac{a_{n+1}}{(n+1)a_n} \\ &= \frac{a_{n+1}}{a_2 \times 3 \times 4 \times ... \times n \times (n+1)} \\&=\frac{a_{n+1}}{(n+1)!} \end{aligned}

Part 3 : Final Step

Suppose P(q)=S(q)P(q) = S(q) for some natural number qq.

ThenP(q+1)=P(q)(1+f(q+1)aq+1)=P(q)+f(q+1)aq+1×aq+1(q+1)!=P(q)+f(q+1)(q+1)!=T(q)+f(q+1)(q+1)!=T(q+1) \begin{aligned} P(q+1) &= P(q)(1 + \frac{f(q+1)}{a_{q+1}}) \\&= P(q) + \frac{f(q+1)}{a_{q+1}}\times\frac{a_{q+1}}{(q+1)!} \\&= P(q) + \frac{f(q+1)}{(q+1)!} \\&= T(q) + \frac{f(q+1)}{(q+1)!} = T(q+1) \end{aligned} as required.

End of Proof

We'll go through some of the consequences of this in the next note!

#Series #TaylorSeries #Summation #Productseries

Note by Roberto Nicolaides
6 years, 4 months 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

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...