Method 3? - Manipulate a power series to look like the designated sum. To start, note that exp(x) = x^0 / 0! + x^1 / 1! + x^2 / 2! + ... From that, we see that exp(-x) = x^0 / 0! - x^1 / 1! + x^2 / 2! - x^3 / 3! ... Adding these together, we get that exp(x) + exp(-x) = 2x^0 / 0! + 2x^2 / 2! + 2x^4 / 4! + ... If we multiply both sides by x/2, we get [x exp(x) + x exp(-x)]/2 = x^1 / 0! + x^3 / 2! + x^5 / 4! + x^7 / 6! + .... If we differentiate both sides, we get [x exp(x) + exp(x) - x exp(-x) + exp(-x)]/2 = 1 x^0 / 0! + 3 x^2 / 2! + 5 x^4 / 4! + 7 x^6 / 6! + ... If we set x = 1, we get exp(1) = 1/0! + 3/2! + 5/4! + 7/6! + ... Note that the right side is the equation we were given. So the sum is equal to exp(1), or e.
Essentially you split the given series into the power series for _sinh(1)_ and _cosh(1)_ respectively, revealing the sum to be *_Σₙ₌₀∞{ (2n+1)/(2n)! } = sinh(1) + cosh(1) = e_* _sinh(1) = 1/1! + 1/3! + 1/5! + ..._ = _2/2! + 4/4! + 6/6! + ..._ _cosh(1) = 1/0! + 1/2! + 1/4! + 1/6! + ..._
The factorial in the denominator makes it obviously converge by the ratio test. On the other hand, it's a good habit to state obvious things and why they're obvious, so that you don't forget that condition when it's not true or not obvious.
@@CriticSimon Beyond infinity does not make sense in math. Infinity is that which has no end. Why don't you try to give an example of something that is beyond infinity. In the case of the series' given in the video, they do not extend beyond the smallest infinite cardinal, aleph-0.
1:48 to be super correct that first 1 should go to bottom row since you can split it into (0+1)/0! so in top row you leave 0/0! and 1/0! goes to the bottom one, i mean doesn't matter, just a quick thought i had
Unlike most of your videos with two (or more) methods, the two methods in this video are exactly the same method but with different notation. That doesn't really count as two methods.
5:42
“Wir mussen wissen. Wir werden wissen. (We must know. We will know.)
[Inscribed on his tomb in Gottingen.]”
― David Hilbert
my small brain method: differentiate the power series of x*cosh(x)
You've got a BIG BRAIN! 😍😍
Did it the same way!
DAMN
Nice!
Method 3? - Manipulate a power series to look like the designated sum.
To start, note that exp(x) = x^0 / 0! + x^1 / 1! + x^2 / 2! + ...
From that, we see that exp(-x) = x^0 / 0! - x^1 / 1! + x^2 / 2! - x^3 / 3! ...
Adding these together, we get that exp(x) + exp(-x) = 2x^0 / 0! + 2x^2 / 2! + 2x^4 / 4! + ...
If we multiply both sides by x/2, we get [x exp(x) + x exp(-x)]/2 = x^1 / 0! + x^3 / 2! + x^5 / 4! + x^7 / 6! + ....
If we differentiate both sides, we get [x exp(x) + exp(x) - x exp(-x) + exp(-x)]/2 = 1 x^0 / 0! + 3 x^2 / 2! + 5 x^4 / 4! + 7 x^6 / 6! + ...
If we set x = 1, we get exp(1) = 1/0! + 3/2! + 5/4! + 7/6! + ...
Note that the right side is the equation we were given. So the sum is equal to exp(1), or e.
Wow! That's cool
Essentially you split the given series into the power series for _sinh(1)_ and _cosh(1)_ respectively, revealing the sum to be
*_Σₙ₌₀∞{ (2n+1)/(2n)! } = sinh(1) + cosh(1) = e_*
_sinh(1) = 1/1! + 1/3! + 1/5! + ..._
= _2/2! + 4/4! + 6/6! + ..._
_cosh(1) = 1/0! + 1/2! + 1/4! + 1/6! + ..._
I think that, first of all, you should prove that the series you're using are not divergent.
The factorial in the denominator makes it obviously converge by the ratio test. On the other hand, it's a good habit to state obvious things and why they're obvious, so that you don't forget that condition when it's not true or not obvious.
You know I generally don't do that! It's too much work and I'm lazy! 😁😁
0:21 "All the way to infinity and beyond". That only makes sense in "Toy Story" movies.
It also makes sense in math. Infinity is not the end...
@@CriticSimon Beyond infinity does not make sense in math. Infinity is that which has no end. Why don't you try to give an example of something that is beyond infinity.
In the case of the series' given in the video, they do not extend beyond the smallest infinite cardinal, aleph-0.
1:48 to be super correct that first 1 should go to bottom row since you can split it into (0+1)/0! so in top row you leave 0/0! and 1/0! goes to the bottom one, i mean doesn't matter, just a quick thought i had
thanks for sharing!
Unlike most of your videos with two (or more) methods, the two methods in this video are exactly the same method but with different notation. That doesn't really count as two methods.
S=Σ(2k+1)/(2k)!=Σ2k/(2k)!+Σ1/(2k)!=Σ1/(2k-1)!(k=1,2,3 ..)+Σ1/(2k)!(k=0,1,2..)=1+1/3!+1/5!+1/7!...+(1+1/2!+1/4!...)=e-1-(1/2!+1/4!+1/6!..)+(1+1/2!+1/4!..)=e-1+1=e