ππ

∑n=0∞((12)n)2(2)n.n!=2F1(12,12;2;1)=1Γ2(32)=4π\sum_{n=0}^∞ \frac{((\frac{1}{2})_n)^2 }{(2)_n .n!} = _2F_1 (\frac{1}{2},\frac{1}{2};2;1)= \frac{1}{\Gamma^2{(\frac{3}{2})}}= \frac{4}{π} ∑n=0∞(23)n(13)n(2)n.n!=2F1(23,13;2;1)=934π\sum_{n=0}^∞ \frac{(\frac{2}{3})_n (\frac{1}{3})_n }{(2)_n .n!}= _2F_1 (\frac{2}{3},\frac{1}{3};2;1) =\frac{9\sqrt{3}}{4π} ∑n=0∞(34)n(14)n(2)n.n!=2F1(34,14;2;1)=823π\sum_{n=0}^∞ \frac{(\frac{3}{4})_n (\frac{1}{4})_n}{(2)_n .n!}= _2F_1(\frac{3}{4} , \frac{1}{4} ;2;1)= \frac{8\sqrt{2}}{3π} ∑n=0∞(45)n(15)n(2)n.n!=2F1(15,45;2;1)=25sin⁡(π5)4π\sum_{n=0}^∞ \frac{(\frac{4}{5})_n (\frac{1}{5})_n }{(2)_n .n!} = _2F_1(\frac{1}{5},\frac{4}{5} ;2;1) = \frac{25\sin{(\frac{π}{5})}}{4π}

And many more...\textrm{And many more...}

In all cases I have used 2F1(a,b;c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)_2F_1 (a,b;c;1)= \frac{\Gamma{(c)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}

I have noticed that , we can easily found out an Infinite sum generator for ππ

When 2F1(1−1a,1a;2;1)=a2sin⁡(πa)(a−1)π\color{#20A900}_2F_1 (1-\frac{1}{a} , \frac{1}{a} ;2;1) = \frac{a^2 \sin(\frac{π}{a})}{(a-1)π} When a=97a=97 then it is 940996πsin⁡(π97)\large\frac{9409}{96π} \sin(\frac{π}{97})

Those Expressions can be transformed into infinite sums . For example

2F1(12,12;2;1)=1+12(12.1!)2+13(1.322.2!)2+14(1.3.523.3!)2+⋯_2F_1(\frac{1}{2},\frac{1}{2};2;1)= 1+\frac{1}{2} (\frac{1}{2.1!})^2+\frac{1}{3} (\frac{1.3}{2^2.2!})^2 +\frac{1}{4}(\frac{1.3.5}{2^3.3!})^2 +\cdots

Proof of Above Identity Take z=1z=1 , then it becomes 2F1(a,b;c,1)=Γ(c)Γ(b)Γ(c−b)∫01tb−1(1−t)c−b−a−1dt_2F_1 (a,b;c,1)= \frac{\Gamma{(c)}}{\Gamma{(b)}\Gamma{(c-b)} } \int_0^1 t^{b-1} (1-t)^{c-b-a-1}dt =Γ(c)Γ(b)Γ(c−b)B(b,c−b−a)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)= \frac{\Gamma{(c)}}{\Gamma{(b)}\Gamma{(c-b)}} \Beta{(b,c-b-a)} = \frac{\Gamma{(c)} \Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}

#Calculus

Note by Dwaipayan Shikari
5 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} ai−1 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 sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Very nice! Put in it Percy's Tau discussion!

Yajat Shamji - 4 months, 4 weeks ago

I will add more values later :)

Dwaipayan Shikari - 5 months ago
×

Problem Loading...

Note Loading...

Set Loading...