@@jamirimaj6880 These would be better presented in a logical , related series so we can build on . or review, that which we have already learned. There is a lot of bringing in items that I dont recall from my days of pure maths A Level.
@@azmath2059 Yes, but his explanations suggest that this is for UNIVERSITY Math beginners. Besides, I have seen many teachers and profs who does these things intentionally, so that in their minds they will see if the students really understand. Sounds good in paper, but these bad habits will affect the flow of the process of explaining rigorous proofs in the long run.
That's cool. I remember working through the "fact" with my daughter a couple of years ago for her 2nd year calculus class. It's a neat thing to have in your toolbox.
Even Putnam-alike or IMC-alike series may be evaluated through telescoping summation! Here we saw that one more crazy series was killed once telescoping summation came into play. Never underestimate the power of telescoping summation!
Yo someone else pointed it out but big oof with the power there, a^(-b) is 1/(a^b) and not a^(1/b) which is the bth root of a. Very big mistake, not sure if it ruins the entire solution
I wonder why when you do partial fractions, you always create a system of equations out of the terms, but there’s an easier way when there are linear terms. In this case, if we plug in u= -1 you cancel the B term and plug in u=1 and you cancel A.
I aggree, just chose some values of the variable. But the method used by Michael is almost doing that for u=0 and then u=1, or, more precisely : first : calculate f(0), and then calculate f(1)-f(0)
Or just prove that 1/(1 - u) + 1/(1 + u) = 2/(1 - u^2). Put everything on the same denominator and you have ((1 + u) + 1 - u))/((1 - u)(1 + u)). Simplify the numerator and distribute the denominator
For the first tool wasn't it simpler just to use difference of squares and subtracting other term? 2/(1-u^2) - 1/(1-u) = 2/(1-u)(1+u) - 1/(1-u) = (2-(1+u))/(1-u)(1+u) = (1-u)/(1-u)(1+u)= 1/(1+u) , without going for coefficients A and B.
Looks fast working! I don't know where I could use this. Maybe if there was some small adjustments it could fit into some problem (not just pure math)...
Dr. Penn putting out those clickbaity titles like @MindYourDecisions. 😁 Merry Christmas, Happy Hanukkah, Seasons Greetings, to Dr. Penn and to all the viewers.
Wolfram alpha couldn’t get an exact value for this in the freely available computation time, it just said it converged to about .44 which is about -1+ 1/ln(2) but computation time had been exceeded. Alpha code: Sum[1/(2^n (1 + 2^(1/(2 n)))), {n, 1, Infinity}] Maybe the pro version could get it exact idk.
The Alpha Code is wrong. In Mathematica it should be: N[Sum[1/(2^n (1 + 2^(1/(2 ^ n)))), {n, 1, Infinity}], 25] which gives 0.4426950408889634073599247 and N[-1 + 1/Log[2], 25] gives the same.
@@coolbikerdad I don’t think the alpha code is wrong, when you put it in it translates it to look exactly like what he has on the board. Do you mean just the 25 for the digit precision?
Unusually I have had a couple of good days. I was able to do yesterday's problem just by looking at the thumbnail and not even starting the video. Today I also did the question but I did start the video and look at the very opening. The thing is having seen the results to be used it is fairly easy to see what is to be done. My self doubt comes from the thought that had I not seen which results were to be used I probably would not have come up with them myself.
True, the few clues he puts are usually enough to solve. But sometimes the thumbnail problem statement is wrong so I usually start the video and hide with my hands the right side and bottom side of the video to see the correct problem statement alone, then start solving it myself.
@@riadsouissi The first result was particularly helpful. Just by seeing it the mind gets wondering why did he suggest that? Then the light comes on and the whole problem became clear. But then I was left with the feeling that I probably would not have taken that approach unprompted. I think the whole problem was constructed backwards by someone who has played around with that particular root of 2.
Would anyone actually thinknofnthat first tool if not given it ir it's just something you have to knkw beforehand..couldnt you solve this by plugging in intial values and trial and error..I think that's how most people would approach it if not given the tools.
ruclips.net/video/Fm5az_95nQs/видео.html There is no need to calculate derivatives in terms of n, we can replace 2^n -> x, so calculating limit x^(-1) / (1 - 2^(1 / x)) is much simpler IMO
Hi, 6:10 : 2^(-x) and not 2^(1/x) For fun: 2 "so let's just go ahead and" 1 "so now let's go ahead and", 1 "we are going to go ahead and" 1 "ok, great", 2 "and so on and so forth", 14:00 "wininin"! 14:46 "the derivative of 1 is 0"
In Germany you say "brackets" but describe which type of bracket: round brackets - runde Klammern ( ) edged brackets - eckige Klammern [ ] curly brackets - geschweifte Klammern { }
@@craftexx15 your comment made me think, since 1 < 2 < e, we know ln(1) < ln(2) < ln(e), so, 0 < ln(2) < 1. That means 1/ln(2) > 1, so I knew Prof Penn's result was a positive number. But I got curious enough to check with the sum term by term in a spreadsheet, and it's indeed correct. The result starts 0.442695..., and you only need about 23 terms in the infinite sum to get 6 decimal places of accuracy.
Despite his beautiful calculation I think that trick is suitable only for one kind of sum. I can't imagine any possible generalization. It's similar to Gaussian integrals Beautiful but there is only small set of integrals that can be calculated that way. I prefer more usable methods ,for example Abel's summation formula or Sommerfeld-Watson transform.
I think the result starts 0.442695..., and you only need about 23 terms in the infinite sum to get 6 decimal places of accuracy. I checked with a spreadsheet.
The little-n chicken scratches! Write bigger and more clearly please -- good god man, some people are older and more feable in the eye but still want to watch your show! Very cool series that is approximately 0.44269504088896340736... to 20 decimal places. Happy Holidays Every One!
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
Nice Infinite sum. Thank you Michael. Merry Christmas.
5:53 note 2^{-x}
eq 2^{1/x} but the original expression on the left of blackboard was correct
I'm convinced each and every single day that he's doing that on purpose, which is actually bad for math beginners who are watching these videos
@@jamirimaj6880 These would be better presented in a logical , related series so we can build on . or review, that which we have already learned. There is a lot of bringing in items that I dont recall from my days of pure maths A Level.
@@jamirimaj6880 This channel is not really for Math beginners
@@highpath4776 This is university level maths
@@azmath2059 Yes, but his explanations suggest that this is for UNIVERSITY Math beginners. Besides, I have seen many teachers and profs who does these things intentionally, so that in their minds they will see if the students really understand. Sounds good in paper, but these bad habits will affect the flow of the process of explaining rigorous proofs in the long run.
**writes (1 - 2ⁿ√2 without closing it multiple times**
"ya that seems like a good place to stop"
this is not a good place to sob.
Those unbalanced parentheses lowkey gave me PTSD. (I'm a programmer).
Must be hard doing what he does, so he sometimes overlooks the little things
That's cool. I remember working through the "fact" with my daughter a couple of years ago for her 2nd year calculus class. It's a neat thing to have in your toolbox.
Even Putnam-alike or IMC-alike series may be evaluated through telescoping summation! Here we saw that one more crazy series was killed once telescoping summation came into play. Never underestimate the power of telescoping summation!
There was no need for second tool at all. Just replace 2^-N with x->0.
"i think that's a good pla-" and the video immediately stops
You can also replace 2^n with 1/x, as x goes to 0. You then are left with x/(1-2^x) which clearly has limit -1/ln2
In 6:02 2^(-x) ≠ 2^(1/x)
yeah, it should be 2^(-x)
I'm guessing they misread their notes. 2^(-x)=1/(2^x)
@@chippetychaps That makes more sense. My brain was screaming when he did that.
Yes he wrote that down incorrectly but the original tool is correct
I thought radical 2 to tbw 2^n power meant the square root of 2 to the nth power..isnt that what it means?
Hats off to you Sir...huge respect...you are one heck-of-a clever dude.
Yo someone else pointed it out but big oof with the power there, a^(-b) is 1/(a^b) and not a^(1/b) which is the bth root of a. Very big mistake, not sure if it ruins the entire solution
the final solution is correct even though he didnt write this down correctly
WHY use the 1 minus u squared integral at all..the form of it is very different from the 1 plus radical n function..so whybuse at all
I like the conviction and gravity with which you articulate. You can be good teacher.
And he never smiles. 😮
NICE - IT CAN BE USED AS PART OF ESTIMATION THEORY - HENCE OR OTHERWISE SHOW THAT EXP(X) CROSSES from 1.xx to 2.xx when x is between 2/3 and 3/4
I wonder why when you do partial fractions, you always create a system of equations out of the terms, but there’s an easier way when there are linear terms.
In this case, if we plug in u= -1 you cancel the B term and plug in u=1 and you cancel A.
I aggree, just chose some values of the variable.
But the method used by Michael is almost doing that for u=0 and then u=1, or, more precisely : first : calculate f(0), and then calculate f(1)-f(0)
i agreed. I mean that systems holds for all u, so whatever u're substituting for u, you can get A and B easier
@@andrijauhari8566 yeah. His method is good to know, because sometimes you have to do it, but for linear terms, it’s overkill.
Or just prove that 1/(1 - u) + 1/(1 + u) = 2/(1 - u^2).
Put everything on the same denominator and you have ((1 + u) + 1 - u))/((1 - u)(1 + u)). Simplify the numerator and distribute the denominator
At about 6:03 I notice a mistake b/c you wrote 2^(-x) as 2^(1/x). But you fixed it in all later steps.
Merry Xmas, professor! Thank you for sharing these.
Great. Happy christmas to you and your family Michael, stay safe!
That was quite informative, Michael. If possible, could you please take some problems involving limits of infinite series?
I have a problem suggestion,
Find all functions f:N-->N such that n!+f(n)+m divisible by f(n)!+f(m+n) for all m,n in N.(0 is not an Natural Number.)
Do you know the solution of this problem or did you just ask it because you think it might be interesting? (To know whether or not I'll try it lol)
14:25 perfect "응"
16:57
This video was uploaded 4 mins ago, you posted this comment 6 mins ago... How the-
@@MelonMediaMedia Time travel
Why the "Good Place To Stop" ritual has been like cuted in the last two videos?
There's that spider/flake of chalk running down the board again 6:25 to 6:28! If it's not put in as an Easter Egg then it should be!
quite a distraction
I saw it!
AHHHHHHH UNCLOSED PARENTHESES PAIN PAIN PAIN PAIN
For the first tool wasn't it simpler just to use difference of squares and subtracting other term? 2/(1-u^2) - 1/(1-u) = 2/(1-u)(1+u) - 1/(1-u) = (2-(1+u))/(1-u)(1+u) = (1-u)/(1-u)(1+u)= 1/(1+u) , without going for coefficients A and B.
Sir I am looking for these kind of crazy integrals, thank you. With ❤️ from India 🎉🎉🎉
Happy Holidays Master
Thank you for your video.
Looks fast working!
I don't know where I could use this. Maybe if there was some small adjustments it could fit into some problem (not just pure math)...
Dr. Penn putting out those clickbaity titles like @MindYourDecisions. 😁
Merry Christmas, Happy Hanukkah, Seasons Greetings, to Dr. Penn and to all the viewers.
Wolfram alpha couldn’t get an exact value for this in the freely available computation time, it just said it converged to about .44 which is about -1+ 1/ln(2) but computation time had been exceeded.
Alpha code:
Sum[1/(2^n (1 + 2^(1/(2 n)))), {n, 1, Infinity}]
Maybe the pro version could get it exact idk.
The Alpha Code is wrong. In Mathematica it should be:
N[Sum[1/(2^n (1 + 2^(1/(2 ^ n)))), {n, 1, Infinity}], 25] which gives 0.4426950408889634073599247
and N[-1 + 1/Log[2], 25] gives the same.
@@coolbikerdad I don’t think the alpha code is wrong, when you put it in it translates it to look exactly like what he has on the board. Do you mean just the 25 for the digit precision?
@@Reliquancy there was a ^ missing. You had a 2n instead of 2^n for the root of 2. And it gave the wrong answer 0.4404.
@@coolbikerdad Oh, I see I couldn’t see that on the chalkboard in the video either on my phone thanks.
Please make a video on Putnam 1999 B4 problem
Unusually I have had a couple of good days. I was able to do yesterday's problem just by looking at the thumbnail and not even starting the video. Today I also did the question but I did start the video and look at the very opening. The thing is having seen the results to be used it is fairly easy to see what is to be done. My self doubt comes from the thought that had I not seen which results were to be used I probably would not have come up with them myself.
True, the few clues he puts are usually enough to solve. But sometimes the thumbnail problem statement is wrong so I usually start the video and hide with my hands the right side and bottom side of the video to see the correct problem statement alone, then start solving it myself.
@@riadsouissi The first result was particularly helpful. Just by seeing it the mind gets wondering why did he suggest that? Then the light comes on and the whole problem became clear. But then I was left with the feeling that I probably would not have taken that approach unprompted. I think the whole problem was constructed backwards by someone who has played around with that particular root of 2.
Hi Michael
Could you please show me why this series doesn’t turn out to be telescoping?
I appreciate your help!
Love these
This feels good watching from Nigeria...
Staggering derivation!
Would anyone actually thinknofnthat first tool if not given it ir it's just something you have to knkw beforehand..couldnt you solve this by plugging in intial values and trial and error..I think that's how most people would approach it if not given the tools.
Merry Christmas
This was cool, thanks.
ruclips.net/video/Fm5az_95nQs/видео.html There is no need to calculate derivatives in terms of n, we can replace 2^n -> x, so calculating limit x^(-1) / (1 - 2^(1 / x)) is much simpler IMO
This series is an infinitesimal part of ramanujam series 😁😁😁😂😉😉😁
8:57 was a good place to stop. Those two sums cancel each other term by term except for n=1 on the left sum (which is the answer)
Let the sum be designated by S
so S < (1/4 +1/4 + 1/8 + .. )
and so S > (1/4 + 1/8 + .. )
so 1/2
No the sum is equal to 0.443 which is less than 0.5
Hi,
6:10 : 2^(-x) and not 2^(1/x)
For fun:
2 "so let's just go ahead and"
1 "so now let's go ahead and",
1 "we are going to go ahead and"
1 "ok, great",
2 "and so on and so forth",
14:00 "wininin"!
14:46 "the derivative of 1 is 0"
Yes he didnt write it down correctly in the derivation, but his original tool is correct
Amazing video. Just love Maths. From Germany
Note that the tools are also facts
So can you simply replace a discrete variable with a continuous one like that?
Only when taking the limit to infinity
I think you cut one more second off the video than you should
superb.
As far as math porn goes...”and that’s a good place to stop” is a bit abrupt.
very nice.
Surprisingly, Mathematica 14.0.0.0 is unable to do this sum.
This sum is so crazy that I thought I was looking at Flammable Maths channel for a moment.
I knew the answer had to be between 0 and 1, at least...
un crack bro
Is there any other countries who say brackets instead of parantheses? In Ireland we say brackets to refer to them.
In Australia we say brackets
In Germany you say "brackets" but describe which type of bracket:
round brackets - runde Klammern ( )
edged brackets - eckige Klammern [ ]
curly brackets - geschweifte Klammern { }
Are you sure that this is right? The result is negative but the summands are all positive. You can not add positive things to get a negative number.
the result isn't negative...
@@kehrierg Oh .. saw it now .. sorry - messed something up 😂
@@craftexx15 your comment made me think, since 1 < 2 < e, we know ln(1) < ln(2) < ln(e), so, 0 < ln(2) < 1. That means 1/ln(2) > 1, so I knew Prof Penn's result was a positive number. But I got curious enough to check with the sum term by term in a spreadsheet, and it's indeed correct. The result starts 0.442695..., and you only need about 23 terms in the infinite sum to get 6 decimal places of accuracy.
good
Nice problem
Nice problem.
to stop.
Ok, ok your click bait worked
Nothing crazy about this infinite series...
I knew I shouldn't click...
I can't see anything crazy about it. Complicated sum leads to complicated nontrivial calculation. That doesn't mean your evaluation isn't perfect.
He has to appease the almighty youtube gods in order to even show up in the recommended videos list of people who are subscribed
Despite his beautiful calculation I think that trick is suitable only for one kind of sum. I can't imagine any possible generalization. It's similar to Gaussian integrals Beautiful but there is only small set of integrals that can be calculated that way. I prefer more usable methods ,for example Abel's summation formula or Sommerfeld-Watson transform.
The decimal expansion of the answer starts -0.30685...
Umm, it can't be negative
I think the result starts 0.442695..., and you only need about 23 terms in the infinite sum to get 6 decimal places of accuracy. I checked with a spreadsheet.
The little-n chicken scratches! Write bigger and more clearly please -- good god man, some people are older and more feable in the eye but still want to watch your show! Very cool series
that is approximately 0.44269504088896340736... to 20 decimal places.
Happy Holidays Every One!
I am the 1000th viewer yay!
so?