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transforming an infinite sum into a finite sum
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- čas přidán 10. 09. 2023
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Fun fact: The identity
a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)
Is also known as the Sophie Germain's identity, named after the French mathematician Marie-Sophie Germain.
was about to right exactly that
She also has a particular set of primes named after her
@@danielprovder after HIM?
@onebronx, @azzteke fixed
btw a^69+420b^666 is named after me
solve the identity urself
that's a very specific, but nonetheless very neat trick. and it's very cool to show that every sum of the form on the lefthand side of the proposition is actually a rational number which you wouldn't necessarily expect from sums like these
Now do the sum of 1/(4n^4+1) and you should get some nice π's and e^π's. :)
Wow!
And no pesky π’s, ln(2), no e^(-1/x).
Thank you, professor.
If any of those values appeared, it would prove that they aren't transcendental numbers. That would be pretty neat if it happened, though.
So it’s actually a telescoping sum in disguise. Neat!
12:09
The satisfaction when that final expression is 1
This trick was completely new for me, as well! Thank you for sharing with us, Prof. Penn!
EXTREMELY clean board.
Another way would be this...
Now sum(1/(n²+(n+1)²) = sum(1/2n²+2n+1)=sum(1/2(n+1)²-2n-1)
Transposing everything other than the first term at n is 2 from 3rd to second equation we get the proposed equation to have the value 1
This video pairs perfectly with @Mathologer recent video on Riemann rearrangement theorem.
Excellent video. I really enjoyed it.
Great video ❤❤❤
Amazing!
that is a nice surprise!
You needed to check that the n=infinity terms go to zero, which they do in this case.
Isn't the original series just a telescoping series? The generalization using the variable a extends the concept of telescoping series from t_n = a_{n+1}-a_n to t_n = a_{n+k}-a_n where k = 2a in this case
Very nice
Could be manipulated so we get zeta(3)? Just shy away from a = 0, but maybe it could be forced somehow. This was my first thought - "we got zeta(3)" then I woke up and stood up from the crapper.
جميل جدا كالعادة
wow!!!
This kinda work for the specific form of that fraction , what if the fraction changes to 5n^2 / 18n^5 +33 ?
Oh but check this out…😂😂😂
Great
Cool!
What happens to the upper limit of the sum if 2a is not a natural number? Maybe I missed something.
We assumed 2a is a natural number right at the very beginning.
Yeh - I fell for that one too
@camilocagliolo Then we wouldn't have been able to do the change of index at around the 8:00 mark.
4:25 can some one explain to me where the 1/(4a) comes from? That is random.
This comes from partial fraction you wil get 2an-(-2an)=4an so you need to divide by 4a to get n.
partial fraction decomposition
we split the initial fraction as a sum of two: A/(n^2 - 2an +2a^2) + B/(n^2 + 2an + 2a^2)
A=-B to cancle out n^2 and 2a^2 terms
and we need 2an*A - 2an*B = n = > 4anA=n = > A = 1/4a
You can kind of guess the PFD: 1/(f-g)-1/(f+g)=2g/(f+g)(f-g). If you match the RHS against the problem, you're only missing a few things from the numerator, so you can just divide by them in front. This comes up all the time when you have a difference of squares in a denominator.
So the question is if all absolutely convergent infinite sums of rational functions can be reduced to infinite sums. The likely answer is no but then what properties does the rational functions have to have so it is true?
It is an interesting thought! Does this example completed by Michael represent a single animal on its own or does it represent a class of things?
Ahhh ... the power of algebraic analysis?
Do you mean reduced to finite sums of rational numbers? If so then this is false for any transcendental. Otherwise it's trivial since you can have the sum of a single value.
@@TheEternalVortex42 I wonder at times if some transformations introduce spurious answers.
Mostly because I don't know enough about compositions of transformations.
As an example .... czcams.com/video/NmJhxa33wnw/video.html
@@Alan-zf2tt There are likely an entire class and on first glance seems to be related to a symmetry in some way. If f,g are rational function with f(x) = g(a+x) - g(a-x) then it the tails of g(x) will cancel.
Basically there is only a finite "middle" part that will contribute to f(x) in the sum as the rest of the tail will cancel itself(it has a type of symmetric periodicity after some point).
Those look like solutions to the wave equation also if g is an odd function so not sure if something is going on in there.
The point is that the function behaves a certain way in it's tail so that terms will cancel out. I think this could get really complex though so I imagine the class is much larger.
Writing an infinite sum as a finite sum obviously shows the infinite sum converges. Wonder if there are infinite sums that could be manipulated like this in which that is the "easiest" way to show that it converges?
I agree. A first thought may be: ahh! Great! Second thought: hey wait a minute! Third thought: get a post-grad in here to check it out!
Fourth though: flippin 'eck - it worked!
👍
Noice!
Here's an obligatory comment asking you how to use math to transform my finite sum of money into an infinite sum of money. Will share $$$ if successful
Oh that's easy! With Banach-Tarski's Theorem it is possible to decompose your money into a finite number of parts such that after reassembling you are left with twice the amount of bills you had. Now repeat the process until satisfied.
Now where are my $$$?
The answer is always something obvious... the maths are optional.