Pi-Squared Over 6 and the Algebraic Genius of Euler - Part I

This is Part I of a 2-part video demonstrating Euler's derivation of a remarkable formula involving pi. This video is for anyone who shares my passion for math and specifically for advanced middle schoolers on up through undergraduate calculus students. Teachers may want to share this with their math club members as well. I apologize in advance for any errors - pls correct them!

Category:

Education

Tags:

• Pi
• Euler
• Math
• Math Education
• Math Enrichment
• Pi Day
• Math Clubs
• Pi-squared over 6

[MathNotationsVids]

dmarain For any function f, the y-int is f(0). For a polynomial, this will be the constant term since all other terms will have a variable. Thus for y=ax^2+bx+c, c is the y-int. Completing the sqr produces y=a(x-h)^2+k. (h,k) locates the vertex of the parabola.

[MathNotationsVids]

Everyone's comments are most appreciated. Thanks for correcting the errors. I'm sure I made others! As far as wasting time, point well taken. I need to tighten up my presentation. I'm not following a tight script.

[gayglue]

at 3:48, don't you have to complete the square for the last term to be the y-intercept?

[johnnynapalmiv]

Wadworth's constant applies

[TOMTOM171717]

seriously, 2 minutes and no maths

[TOMTOM171717]

i think you should dive straight in to the math, people who don't like maths can work that out for themselves

[StefanoCiampichetti]

Also in the second video a little mistak..it should be:

Also in this video a little mistake: 1 - (x^2)/6 + (x^4)/120 - ..... (So, not x^5)

[StefanoCiampichetti]

Hey my friend! Good vid! But you made a mistake: It is: 1 - 1/6x + 5/6x^2

[TLMCiSkyzz]

first post. woot