The product of local extrema of cubic function serves as the discriminant of a cubic equation. If the product is positive, that means the equation has a only one real root; if it is negative, the equation has three real roots; if zero it has a one real root and a multiple one. Here is examples: def f(x)=x³-3x²+5 local maximum == 5 local minimum == 1 product == 5 thus f(x) has a one real root. def g(x)=x³-6x²+9x-1 local maximum == 3 local minimum == -1 product == -3 thus g(x) has three real roots. def h(x)=x³-12x²+36x local maximum == 32 local minimum == 0 product == 0 thus h(x) has a one real root and a multiple real root.
multiply denominators with 2 it collapses into 1 then multiply numerators with 2, thus 2(1) =2. Similar technic with the sum of 1/(k^2)-1, except multiply the numerators first with 2, it collapses into 1+1/2 then multiply with 1/2 thus=3/4.
Method 2. Start with the Taylor series for 2/(1-x). Integrate the series twice. This series will be equal to 2 (x - (-1 + x) log(1 - x)) in the radius of convergence |x|
ATTEMPT: surely there must be a way to telescope by partial fractions this. comparison test is telling me that this is between (pi^2)/3 (by 2 + 2/4 + 2/9 + 2/16 + ...) and (pi^2)/3 - 2 (2/4 + 2/9 + 2/16 + 2/25 + ...). So it definitely converges sum(1/sum(j: j in 1 to i): i in 1 to n) = sum(2/(i(i+1)): i in 1 to n) = 2 * sum(1/(i(i+1)): i in 1 to n) = 2 * sum((i + 1 - i)/(i(i+1)): i in 1 to n) = 2 * sum(1/i - 1/(i + 1): i in 1 to n) = 2 * (1 - 1/(n + 1)) = 2 - 2/(n + 1) Take the limit as n -> +inf: = 2 - 2/(+inf + 1) = 2 - 0 = 2
Enjoyed the presentation, except for one fatal flaw. A warning: it's dangerous to call Ramanujan "wrong". He was (with Hardy and Littlewood) infinitely (!) smarter than you and me both. I know, that with suitable definitions of what "summable" means, indeed Sum[n=1 to inf](n)=-1/12. You, sir, are wrong on this point.
Thanks! Sorry for bursting a bubble: He can be super smart but that's a mistake! The sum is incorrect! He was wrong! (maybe he was just messing around)
Please consider making a video on "The Discriminant of a cubic equation", I searched for it but couldn't seem to find any good ones
The product of local extrema of cubic function serves as the discriminant of a cubic equation.
If the product is positive, that means the equation has a only one real root; if it is negative, the equation has three real roots; if zero it has a one real root and a multiple one.
Here is examples:
def f(x)=x³-3x²+5
local maximum == 5
local minimum == 1
product == 5
thus f(x) has a one real root.
def g(x)=x³-6x²+9x-1
local maximum == 3
local minimum == -1
product == -3
thus g(x) has three real roots.
def h(x)=x³-12x²+36x
local maximum == 32
local minimum == 0
product == 0
thus h(x) has a one real root and a multiple real root.
@@YeonJe-1128what if there are no local extrema?
from Morocco thank you for this clear and complete proof
multiply denominators with 2 it collapses into 1 then multiply numerators with 2, thus 2(1) =2. Similar technic with the sum of 1/(k^2)-1, except multiply the numerators first with 2, it collapses into 1+1/2 then multiply with 1/2 thus=3/4.
k th term = 2 /( k ( k +1))
= 2 /k - 2/( k + 1)
Hereby this telescopic sum equals to
2 /1 -2 /2 + 2/2 - 2 /3 + .. = 2
I didn't notice the(+....) before and was very proud that i was able to solve the and get 1.5 😂😂😂😂
🤣🤣
2 final answer
Exactly the method I used
Génial ❤
👍
Method 2.
Start with the Taylor series for 2/(1-x).
Integrate the series twice.
This series will be equal to 2 (x - (-1 + x) log(1 - x)) in the radius of convergence |x|
How would you get the thought to even use the Taylor series of that function?
What is the solution?
2
1/3 + 1/6 = 1/2 not 5/6
Why to use partial fraction
So that you can split it into 2 sums and see it's telescoping.
Let the nth term is
Sn=1/(1+2+3+...+n)
=1/[½n(n+1)]
=2/[n(n+1)]
=2×[1/{(1/n)-{1/(n+1)}]
1+[1/(1+2)]+[1/(1+2+3)]+...
=2[(1-½)+(½-⅓)+(⅓-¼)+(¼-⅕)+...]
=2
In _my_ math, 1 + ¹/₃ + ¹/₆ = 1 + ³/₆ = 1½, not 1⁵/₆ . Not that that invalidates the rest.
1 + ¹/₃ + ¹/₆ + ¹/₁₂ = 1⁷/₁₂ , 1 + ¹/₃ + ¹/₆ + ¹/₁₂ + ¹/₂₄ = 1⁵/₈ , etc.
The fourth term is 1/10, mr.Know-it-all
@@leonidfedyakov366 Arggh, yes, sum not product. [*sigh*]
sum[k=1,n](k)=n(n+1)/2
S=sum[n=1,♾️](2/n(n+1))
2/n(n+1)=2/n-2/(n+1)
this will telescope, leaving only the first and last terms.
S=2/1-2/♾️
S=2
2
2 ???
lim 2
Routine classroom problem
😊😊😊
S=Σ(n=1...inf)1/n(n+1)/2=2Σ(1/n-1/n+1)=2*1=2
You are missing grouping symbols: 1/[n(n + 1)/2], for starters.
1+1/(1+2)+1+1/(1+2+3)+... = ?
Sum n 1->inf 2/(n^2+n) =
Sum n 1->inf 2/n - 2/(n+1)
2/1-2/(1+1)+
2/2-2/(2+1)+
2/3-2/(3+1) = 2/1
everything cancels except the first.
1+1/(1+2)+1+1/(1+2+3)+... = 2
First Gauss's teacher was a man.
1 + 2 + 3 + • • • n =
(n + 1) + (n - 1) + 2 +
(n - 2) + 3 + • • • = (n + 1) + (n + 1) + (n + 1) + • • •.
I did not know! Thanks 😀
ATTEMPT:
surely there must be a way to telescope by partial fractions this.
comparison test is telling me that this is between (pi^2)/3 (by 2 + 2/4 + 2/9 + 2/16 + ...) and (pi^2)/3 - 2 (2/4 + 2/9 + 2/16 + 2/25 + ...). So it definitely converges
sum(1/sum(j: j in 1 to i): i in 1 to n)
= sum(2/(i(i+1)): i in 1 to n)
= 2 * sum(1/(i(i+1)): i in 1 to n)
= 2 * sum((i + 1 - i)/(i(i+1)): i in 1 to n)
= 2 * sum(1/i - 1/(i + 1): i in 1 to n)
= 2 * (1 - 1/(n + 1))
= 2 - 2/(n + 1)
Take the limit as n -> +inf:
= 2 - 2/(+inf + 1)
= 2 - 0
= 2
the sum is 3
∑[k=1, ∞]2/k(k+1)
= 2∑[k=1, ∞](1/k-1/(k+1))
= lim[k→∞]2(1- 1/(k+1))
= 2
1+1/2(1/2)=5/4
easy
It’s obviously -12…😂
God help us.
Enjoyed the presentation, except for one fatal flaw. A warning: it's dangerous to call Ramanujan "wrong". He was (with Hardy and Littlewood) infinitely (!) smarter than you and me both. I know, that with suitable definitions of what "summable" means, indeed Sum[n=1 to inf](n)=-1/12. You, sir, are wrong on this point.
Thanks! Sorry for bursting a bubble:
He can be super smart but that's a mistake! The sum is incorrect! He was wrong! (maybe he was just messing around)
6:32 i think the notation is wrong, it should be
∞
∑ ( 1 / ∑ⁿᵢ₌₀ i )
ₙ₌₁