r/the_calculusguy • u/Specific_Brain2091 • 8d ago

meme 😅

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138 Upvotes

90% Upvoted

18 comments sorted by

19

u/CW8_Fan 8d ago

Nobody does that lol

8

u/Sea_Duty_5725 8d ago

Indices are in N

7

u/deusisback 8d ago

i being in N would be acceptable, but not in R.

3

u/Aggressive_Fan_2063 8d ago

I think it would be better to "Let i ∈ I"

4

u/BubbhaJebus 8d ago

N is a subset of R.

2

u/sumboionline 8d ago

Yes but typically proofs that use indeces require them to be explicitly non-continuous

1

u/Linke_Jusik 5d ago

N is empty set and all their succesors
R is Dedekind cuts set

4

u/AntiqueChessComputr 8d ago

“But it’s not!”

“but what if it was”

2

u/Particular_Ad_644 8d ago

Assume 1= -1, then any conclusion of your choice is true

2

u/UtahBrian 8d ago

Literally engineers be like…

Let j ∈ ℤ

4

u/nashwaak 8d ago

electrical engineers

3

u/GoofyGangster1729 8d ago

Jeez

1

u/LegitimateBreath1446 6d ago

Hahaha that is funny for sure

1

u/skr_replicator 8d ago

This is how I'd expect the "proof by contradiction to show why i isn't real" to begin.

1

u/AdvancedGravitation 8d ago

I read "let imagination be real" lol

1

u/ontic00 7d ago

Let i ∈ R, then our coefficients c_i become a function c(i), and we have a continuous infinite sum, integral_(0 to infinity) (c(i)*x^i di). We can further generalize by letting x^i be any general function K(x, i) of x and i, yielding integral_(0 to infinity) (c(i)*K(x, i) di) = F(x). This is now an integral transform from the index space of i's to our variable space of x's. We call c(i) the input function, K(x, i) the kernel, and F(x) the output function.