sum versus product battle!!

[@emanuellandeholm5657]

Relationships between sums and products are really interesting. Thanks for this video professor Penn!

[@vilkillian]

yo. Formula on the thumbnail is wrong. As i saw that i thought that tan(m*pi) where m belongs to Z is always zero, so product and sum is 0. nothing to compare But in video you have 1/(2n+1) inside tan argument, not outside like in the thumbnail

[@adityamohan7366]

Funny how the product is less than the sum for n > 1. The sum is exactly (Product) x (Product - 1) /2

[@rajballabhyadav5089]

An alternate way could be to use the relation tan(a+b+c+d.......)=(s1-s3+s5-s7+......)/(1-s2+s4-s6+.....), you can find the formula on web and get to know what the s's are. According to question we will set a b c d..... (there are 2n+1 letters in the argument of tan) equal to mπ/2n+1 and hence the above equation gives 0 on LHS and implies that the numereator be 0 , this gives a polynomial in tan^2(mπ/2n+1) after which the results could be derived. It may look lengthy(just because I have written a lot of things) but I found it easier and calcultion(further steps) are super short using vieta's relations. NOTE: the polynomial should be considered for two different cases i.e. when n is odd and when it's even, it wont' change the results but will only change the signs of coefficients in the polynomial. Also the thumbnail expressions are misleading.

[@AnshPr2311]

More easier is: (mπ)=[(2n+1)∅] now take tan on both sides.. Numerator of expansion of RHS =0 now you'll get the polynomial of increasing odd powers of tan,but it'll give us the roots tan(mπ/2n+1) but we need tan^2 hence we can use transformation of roots then we'll use vieta's 🙏🏻

[@dcterr1]

Wow, impressive calculations!

[@blackflan]

So interesting how the product of the tangents is equal to the mean of them (the sum divided by n). Is there some connection to some probability distribution?

[@Mephisto707]

Imagine having to figure all this out in a test!

[@JirenSlr]

I actually got these exact sum and product to work out in a test on my first year of a college

[@kucckumelon2837]

​@@JirenSlrwhich class was this asked in?

[@JirenSlr]

@@kucckumelon2837 I believe it was analysis

[@morelelfrancel6603]

This is beautiful.

[@gabrieltellesm]

Amazing!

[@francoismichelon2848]

Maybe you should be a bit more careful when you divide by cos(m*pi/(2n+1)) and check when it is equal to 0

[@michaelempeigne3519]

integrate [ (e^x ) * ( cos x ) * ( ln x ) dx ]

[@housamkak646]

insaneeeeee loveed itt

[@chopinmin]

Great video! Just a quick comment that the expressions in the thumbnail are not quite right...!

[@thephysicistcuber175]

Unforseen solution. This is so cool.

[@VishnuSrivatsava]

Hey you're into math too? Nice! No more uploads?

[@thayanithirk1784]

Please do some geometry problems and also try IIT JEE ADVANCE questions they are really cool

[@kamalnehra4295]

Jee advanced questions are cool but not much intuitive . There is a lot of calculation and less thinking

[@V-for-Vendetta01]

@@kamalnehra4295 what are you smoking?

[@kizyzo1348]

@@kamalnehra4295 have you ever actually tried a that paper for yourself atleast once?

[@gmchess7367]

Mind blowing question 👍👍

[@karimjemel7405]

How can you divide inside the sum when the term you're dividing with is non constant (in terms of k)?

[@movax20h]

Vieta's formula in disguise.

[@michuosas]

hope I will someday get to work this way.

[@warmpianist]

I have some sort of weird question out of my mind. What are all possible real functions f such that sum f(m) from m=1 to n is equal to product f(m) from m=1 to n, for all n?

[@demenion3521]

you can transform the equality to ln(sum(f(m)))=sum(ln(f(m))). for n=2, there is a family of solutions, but I think for n>2, you quickly run into problems because of the concavity of the logarithm.

[@odenpetersen6028]

"Elementary techniques"

[@electroskylightgaming4085]

Shut up

[@santiagoarce5672]

What part would you consider non-elementary? I think maybe the only thing not taught in high school is the pi sign but that is pretty easy to understand.

[@JoaoVictor-gy3bk]

They're elementary, they're just being used in a more "complicated" way, but are elementary nonetheless

[@goodplacetostop2973]

17:31

[@santiagoarce5672]

Who are you?

[@nasim09021975]

I see these videos sometimes have glimpses of Deja Vu (repetition of sentences) 😄

[@tianyouli9762]

thanks！

[@angelgabrielramirez9494]

My man over here killing my self esteem

[@bartolhrg7609]

Just another proof that mathematicians choose a few formulas Mix them up A expect you to guess them during the test

[@kqp1998gyy]

AweSome!🌷

[@williamadams137]

7:34 and that’s all equal to zz...

[@famoxyzfamoxyz7027]

1:05 i m pi. that's irrational of michael to say.

[@elhassananhichem2677]

Think for alle .please i need your wathsup proof .

[@zygoloid]

Early on, you rewote e^imπ as (e^imπ/x)^x. I would have liked to see more justification of this step, since it's not correct in general. For example, if x = ½ and m = 1, then the LHS is -1 but the RHS is 1^½ = ±1. I think it probably is correct when x is an integer (as is the case here), but it would have been useful to me if you'd talked through that subtlety a little.

[@otakurocklee]

good point!

[@otaibabahar7521]

Zeta (i) =?

[@nevokrien95]

The thumbnail is missleading

[@FisicoNuclearCuantico]

:P I work for the U.S. govt. I've got to stay lowkey.