Hi sci, I bet you can't tell what went wrong. PIC related

>>7765541
Termen =terms
Differentieren = derivative

>>7765541
You forgot to differentiate "n terms" which is another n-dependency.

i like how from 2n=n you were able to conclude 2=1 rather than n=0

>>7765547
The initial statement holds for all n.

x=5
take derivative of both sides
1=0

hurr durr

good thread

>>7765559
good bump

Professor approved

>>7765561
>underlining vectors
wew

>>7765561
The book uses bolded symbols for vectors

>>7765541
confusion of differentiation of a discrete variable with a real valued continuous one, implied by "n terms" which is not possible for non natural number values

[eqn] n^2 = \underbrace{n + n+ \dots + n}_{n~ \text{termen}} [/eqn]
differentieren
[eqn]2 n = \lim_{h \to 0} \frac{\underbrace{(n+h) + (n+h) + \dots + (n+h)}_{n+h~ \text{termen}} - \underbrace{n + n+ \dots + n}_{n~ \text{termen}}}{h}[/eqn]
[eqn]= \lim_{h \to 0} \left( \frac{\underbrace{h + h + \dots + h}_{n~ \text{termen}}}{h} + \frac{\underbrace{(n+h) + (n+h) + \dots + (n+h)}_{h~ \text{termen}}}{h} \right)[/eqn]
[eqn]= \lim_{h \to 0} \left(\underbrace{1 + 1 + \dots + 1}_{n~ \text{termen}} + \frac{h (n+h)}{h} \right)[/eqn]
[eqn] = \lim_{h \to 0} \left(n + (n+h) \right) [/eqn]
[eqn] = 2n [/eqn]

>>7765583
>(n+h)+...-n+n+...

>>7765549
so ?

>>7765541
n^2 = 1+1+...+1 (n^2 times).
Differentiate:
2n = 0.
n=1:
2 = 0.

>>7765596
So you can't arrive at a restriction on n.