Series solved with increasingly advanced mathematics
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- Опубліковано 28 чер 2024
- What is the sum of the infinite series n squared divided by 2 to the n? I present 3 different solutions!
Yes this is a re-upload. @Marco12388 alerted me to a typo in the original upload around 2:20 where I skipped the 16 denominator by mistake. Original video: • Series solved with inc...
0:00 problem
0:42 important step
1:35 high school
3:16 college
7:03 genius level
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As a college grad who can not solve this question, I will enroll in middle school next year.
What did you study though, if I may ask?
honestly i near fell asleep by the time i got past all the ads, but i'm glad i did -- excellent explain, thank you
How many ads do you have?
Why don’t you use an ad blocker?
Whenever an ad appears just close the video . Then open it again if ad reappears then again close the video. Repeat this process about 3-4 times.
Believe me, UA-cam Premium is worth far, far more than the twelve bucks a month it costs. No ads, no muss, no fuss.
an arithmo-geometric series is just the sum of multiple geometric series. and that sum is a geometric series. that is truly awesome.
I didn't know that either. I learnt something.
Yes this is a re-upload but had to fix a typo. @Marco12388 alerted me to a typo in the original upload around 2:20 where I skipped the 16 denominator by mistake. Original video: ua-cam.com/video/gBLf0o6nzMQ/v-deo.html
I thought I was experiencing a glitch in the matrix lol
at 6:08 you said "a common ratio of 1/4". i think it should be 1/2.
@@sutapadey5274time for a re-re-upload lol
Great video. I always struggled with series and sequences throughout my electrical engineering degree. Please do some more!
Very interesting! I have never had much experience trying to find the value of a series because I guess it was never relevant to my study. I agree to watch carefully for absolute convergence before you do things with arithmetic. You didn't actually mention the absolute part, but since it's a positive series, convergence is sufficient. I think I'll look online to find some practice problems to try some of these ideas. They aren't hard concepts, mostly trial and error, but I've just never done it.
For some reason I was able to understand the college solution better than the high school one, you have an amazing way of explaining problems that most people would initially view with fear, it's great how you introduce multiple ways of approaching such problems so that people of different learning styles can look at different perceptions, keep up the good work! ✨✨✨✨
Very enjoyable! Thanks
Brilliant! So, where can we find more foundation information about sequences and series?
Jee advanced notes
At 3:48 it's stated that the n=0 term is 0 (when it looks suspiciously like 1/2) and goes on to prove at 5:01 that both sum(n/2^n) and sum(1/2^n) are 2; which is fabulous (literally) since if you subtract the latter from the former you get something that's decidedly positive and not infinitesimal. Fun stuff!
I think the n=0 term which is 0, he was referring to is the right hand side term. Basically by adding 0 to the left hand side you can remove the shift from n to n+1.
i was wondering about the sum(n/2^n)=sum(1/2^n) business aswell. on wolfram alpha it says both are 2
Now , This is the content we want ❤❤❤❤
I solved it using some sort of a combination of the "high school" approach and the "college" approach. The "PhD Level" approach is quite impressive, though! "Generating functions", I must remember that.
I like the college level the most, simple and elegant
the 2nd solution was beautiful
did not see that coming
Very cool to see multiple methods for the same problem. While Method 3 was labeled the “genius level” method, it did seem like it required the least amount of creativity. All the manipulations are fairly straightforward ones you see with generating functions. If you were to use Method 3, are there additional justifications you would have to make? I tend to forget when you have to actually prove the radius of convergence
That justification was made at the start of the solution. The series 1+x+x^2+x^3+… always converges when |x| < 1
Specifically, it always converges to 1/(1-x), though I can’t recall how the proof of that fact goes
@@frimi8593
S = 1 + x + x^2 + ... + x^n
xS = x + x^2 + x^3 + ... + x^(n+1)
S - xS = 1 - x^(n+1)
(1 - x)S = 1 - x^(n+1)
S = (1 - x^(n+1)) / (1 - x)
As n -> infinity, x^(n + 1) -> 0, under the assumption that |x| < 1
Therefore, S -> 1 / (1 - x)
You could have calculated the sum n/2^n with the same trick you used for the sum n^2/2^n, i.e.
green sum = sum_{n=0}^\infinity n/2^n, blue sum = sum_{n=0}^\infinity (n+1)/2^(n+1)
and then sum = 2 * blue sum - green sum, which leads to a geometric series
The final method using calculus is awesome
Infinite series questions have always caused me a giant mental block. Still trying to figure out the first problem given to me ......
infinite sums like these (polynomial times an exponential) are always rational, and i'm pretty sure there's an upper bound for the denominator of the answer. the 100th partial sum of this one's continued fraction is [5; 1, 121819200483204824283750066, ...], and the third term is definitely above the upper bound, meaning the answer is 5 + 1/1 = 6
I'm not even sure I could have figured out the high school method on my own. The best I'd do is say it converges.
Really impressive...i learned these methods ...in class 12th ..
I did it similarly to the third approach but slightly differently. My guess was that the n2 made me think of the second derivative of a geometric series.
Like in the video I defined f(x) = sum(x^n, 0, +inf) = 1/(1-x), and S(x) = sum(n^2 * x^n, 0, +inf)
I differentiate once and get :
sum((n+1)x^n, 0, +inf) = 1/(1-x)^2
And twice :
sum((n+2)(n+1)x^n, 0, +inf) = 2/(1-x)^3
Then I expanded the term of the second derivative to express it in terms of n^2, the first derivative and a constant. I get
(n+2)(n+1) = n^2 + 3(n+1) - 1
Now I can rewrite the second derivative in term of S(x), f’(x) and f(x) :
f’’ = S + 3 f’ - f
You solve for S and fix x=1/2 to get the answer.
(For those wondering, I am not a mathematician, just a French engineer)
That's almost how I did it, but I got
S = x*f'(x) + x^2*f''(x)
for x = 1/2.
@@juttagut3695 oh I see. You did not reindexed the derivative series, don’t you ? I didn’t do that by fear of forgetting the first terms. But I guess it works too ! 👍
Three for the price of one. A Black Friday special!
In India, we actually use the 3rd level technique to solve problems. We also use integration in some sequences.
Yes...
But I prefer summation one only.
That's better in 90% cases
@@epikherolol8189 true. The other cases are only used for specific scenarios where summation is useless. MindYourDecisions just wanted to educate us on some potential approaches we may need in future. So it is a noteworthy technique.
whether the numerator is 1 or N being inconsequential is hard to grasp. Seems like n/2^n > 1/2^n for all positive n > 1
Is there another “best community on UA-cam” where they show this graphically? Mathematical Visual Proofs, 3Blue1Brown, Digital Genius, Mathologer… those kinda higher level channels?
Series will be the bane of my GPA
Generating functions way is also olympiad level
Good video thank you
There is a very useful idea that $\frac{1}{(1 - x)^m} = \sum\limits_{n = 0}^{\infty}\binom{m + n - 1}{m - 1}x^n$, which can be used extensively here.
I'm not even at this video's high school level.
My knowledge of math is like my piano playing. I have a few basics down and understood, and can play a few pretty numbers, but I'm not really that good at it. It just can appear I am, because I have some solid fundamentals learned, and know the theory behind it. Though, in practice, like in music, I go off rhythm. It's more a proficiency at certain VERY BASIC theory concepts.
But I can understand other people, and the processes they use. Just like when I watch an expert play Piano, I can see what they're doing, but can't replicate it. So I can follow the logic in most of these videos. I just can't do it, and that because of a writing disability that causes me to get things ordered backward. Which is why I have trouble with minuses when doing very basic math.
Don't worry, even at the high school level explanation, he uses calculus to justify he can even do it. This topic isn't really relevant for high school before calculus. Interesting comparison to the piano by the way.
@@mike1024. Thanks. I understand what he's doing. I just can't do this kind of math.
Yeah I wondered it when I was in high school. Calculus is OP, but this is simply so clever
I hope by 'highschool' he did not mean this was a standard problem but rather a competition based problem because it took me a while to get it haha.
The last one really is something😂
💀I genuinely solved it by the last method. We have been taught this type in entrance exam of jee advanced
sir u mean that we can find any summation by derivative?
I found the phd level easier than the school level
The school level proof was too hard 😢😢
Yeah school lvl was not intuitive at all
6:06 isn't the ratio 1/2 and not 1/4?
and he's gonna have to reupload the video
Multiplying by 2 is a COMPLETELY unnecessary step and only makes the problem more difficult than it needs to be
i used summaton by parts
The phd level approach is imho the easiest one.
Can anyone put this fraction into a computers program ? Can you make an algorith to solve this type of math problems ? :p
I think I can give a fourth, more perfect method
Me as a coms student just make code of this series and calculate the close value
I'm a math student and i also learnt coding so yes. I wouldn't think of manipulating the power series of the geometric series though, or using the other 2 methods. I do recognise instantly that I can use the ratio test to see if the series converges
I can only think of manipulating the series cause I am still in high school. But other two methods are really impressive.
The phd level is the only way to reach God mode🗿✅✅
🎉🎉🎉🎉🎉🎉🎉
I feel like I just got whooped
The math really is not all that advanced. Any math major should have no problem with this.
ez
*Thumbs-down!* That is a relative statement, and you were not wise in stating that.
@@robertveith6383 aww so sad. im a college student n this is freshman year stuff, im a junior. yes its relative. but holds true