An Ultraconvex Putnam Problem.

Taylor series identity: • Proof of the Exact Err...
2018 Putnam A5. This Putnam competition problem brings together a ton of different ideas in calculus in a very cool way! This is one of my favorite competition problems.
Check out more calculus problems: • Calculus Problems
New math videos every Monday and Friday. Subscribe to make sure you see them!
Music: C418 - Pr Department

[@KinataKnight]

I was excited to see this problem was on the Putnam because I asked essentially the same question on MSE a few years ago! (If you're curious, the question is "Can a smooth function have all derivatives positive for x>0 and 0 at x=0? "). Alas I'm a grad student, so this did not occur when I was actually taking the Putnam myself.

[@stefanosvasileiadis2732]

Very nice proof with use of simple mathematical tools. Totally liked this video!

[@masoomladka8017]

Full calculus revised hehe😁😂

[@Stormneo]

I was here before 1K subs, remember me when you blow up

[@Stormneo]

Update: you’re blowing up

[@NeerajKumar-gk9kz]

Hi can u tell me what resources study before the solve this putnam problam how can i solved this type problam putnam plz suggest me

[@mathematicalvillage2357]

This is really nice

[@xaxuser5033]

plzz try this problem : consider f(x)=3x^2+1 . prove that the product from i=1 to i=n of f(i) has at most n distinct prime divisors

[@rohinivankudre5671]

Very nice video. This can be used for all calculus probs related to functions! What are you currently doing? (Meaning which grade/ undergrad yr/ etc are u in)

[@MuPrimeMath]

I just graduated high school!

[@santiagoarce5672]

Where did you get the epsilon delta shirt?

[@MuPrimeMath]

I got it from blackpenredpen: teespring.com/stores/blackpenredpen

[@santiagoarce5672]

@@MuPrimeMath Oh dang he doesn't sell it anymore. They only have the hoodie version now. Thanks for the swift response though!

[@bigjazbo9217]

He used u = 1, but that means x = 0, whereas the condition was x > 0. So shouldn't he evaluate f(u) = f(1) as limit as u approaches 1+? He skipped this step.

[@MuPrimeMath]

The substitution was u = x - 1. This is equivalent to x = u + 1. Therefore, if u = 1, that means x = 1 + 1 = 2, not 0. Also remember that at 21:55, I switch the u back to an x without reversing the substitution because u is a dummy variable.

[@yuitocheng8919]

So it means f(x) is simply the zero function because 1 is not special, right?

[@MuPrimeMath]

Yes, any ultraconvex function with f(0) = 0 must be zero everywhere!

[@vinayaklahoti]

You hit randomly at a triangular dashboard with sides/edges in Geometric Progression of a common ratio of 1.1. You win when the ratio of distance between point of strike and orthocentre and distance between point of strike and incentre is less than 1.1 What's your probability of winning? Can you solve this please?