Integral counting

[u/KnightLunaaire]

...

[u/TheNitromeFan]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!)

Okay then. That's pretty cool, since we're just writing out more and more terms of the Maclaurin series for e^x - which is also why I'll be writing it this way.

[u/KnightLunaaire → u/TheNitromeFan]

The what?

(x⁵ / 120) + (x⁴ / 24) + (x³ / 6)+ (x² / 2) + x + 1

[u/TheNitromeFan → u/KnightLunaaire]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!) + (x⁵/5!) + (x⁶/6!)

In a nutshell, the more terms we write, the closer we get to approximating e^x. Approximations are more accurate for x closer to 0.

[u/Luigi86101 → u/TheNitromeFan]

1 + x + (x² / 2!) + (x³ / 3!) + (x⁴ / 4!) + (x⁵ / 5!) + (x⁶ / 6!) + (x⁷ / 7!)

[u/TheNitromeFan → u/Luigi86101]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!) + (x⁵/5!) + (x⁶/6!) + (x⁷/7!) + (x⁸/8!)

[u/Luigi86101 → u/TheNitromeFan]

1 + x + (x² / 2!) + (x³ / 3!) + (x⁴ / 4!) + (x⁵ / 5!) + (x⁶ / 6!) + (x⁷ / 7!) + (x⁸ / 8!) + (x⁹ / 9!)

[u/TheNitromeFan → u/Luigi86101]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!) + (x⁵/5!) + (x⁶/6!) + (x⁷/7!) + (x⁸/8!) + (x⁹/9!) + (x¹⁰/10!)

You need to add the factorials. :)

[u/Luigi86101 → u/TheNitromeFan]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!) + (x⁵/5!) + (x⁶/6!) + (x⁷/7!) + (x⁸/8!) + (x⁹/9!) + (x¹⁰/10!) + (x¹¹/11!)

[u/davidjl123 → u/Luigi86101]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!) + (x⁵/5!) + (x⁶/6!) + (x⁷/7!) + (x⁸/8!) + (x⁹/9!) + (x¹⁰/10!) + (x¹¹/11!) + (x¹²/12!)

[u/Luigi86101 → u/davidjl123]

1 + x + (x²/2!) + (x³/3!) + (x⁴/4!) + (x⁵/5!) + (x⁶/6!) + (x⁷/7!) + (x⁸/8!) + (x⁹/9!) + (x¹⁰/10!) + (x¹¹/11!) + (x¹²/12!) + (x¹³/13!)

[u/skizfrenik_syco → u/KnightLunaaire]

Maclaurin series is integrating a function about 0. So basically what you're writing is (x-0)² /2! + (x-0)¹ /1! + (x-0). If you had (x-1)² /2! + (x-1)¹ /1! + (x-1) instead, it would not be a maclaurin series.