a nice 'advanced' calculus problem

We look at a nice application of the intermediate value theorem.

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[MichaelPennMath]

Speaking of Calculus... Make sure and tune-in for the epic Calculus livestream this Sunday (June 27) starting at 8am EDT. I will be covering an entire course in integral calculus!

[HAL-ojjb]

Mathematician: this is an incredibly rough sketch of the proper proof
Me, a physicist: hey, what a rigorous proof!

[paolopiccione]

Thanks Michael. Just a minor point: it is g(1)<3/2<g(0).

[tavishu]

I thought the problem was to actually solve the equation for t, and spent one full hour trying to do that in vain 😮. Managed to show that 1/8 < t < 1/4.

[sergeyd]

At the 4:15 mark Michael writes g(0)<3/2<g(1), what is wrong, because above he wrote the opposite. He repeats that at 9:54 again.
At 9:27 he writes g(t)-g(t) while saying "g(t)-g(a)".
Those are just normal mistakes which do not change the solution process, but I was glad I spotted them.

[andyneeman]

4:23 how is g(0) < 3/2 < g(1)? g(0) is 1.718 which is bigger then 3/2, and g(1) is 1 which is smaller then 3/2. Shouldn’t the order of the inequality be reversed? g(0) > 3/2 > g(1)?

[tomasstride]

This is one which is very clear what to do having seen the initial hint.

[theash]

Alternatively, the "easy" way to show that g(t) is continuous is to use linear algebra. Since x^t and e^x are both continuous functions, and the field of continuous functions is closed under multiplication, x^t e^x is continuous. And since integration over an interval is an isomorphism in the vector field of continuous functions g(t) is continuous. The math is simpler, but relies on linear algebra that's past the calculus used for this problem (although which is fairly intuitive).

[goodplacetostop]

HOMEWORK : Find all finite sets M of real numbers such that, whenever a number x is in M, then the number x² - 3|x| + 4 is also in M.

SOURCE : 2016 Alberta High School Mathematics Competition

[anmoldeepsingh]

Could we have used the maclaurin expansion of e^x, then integrate the generated polynomial, then interpret the resulting function in t as a sum of multiple hyperbolic function on the graph and use its continuity in (-1, 1) and value at t=0 to prove we have a solution for t in (0, 1)

[abdullahyousef]

4:40 isn't g(0) supposed to be greater than 1.5 and g(1) less than 1.5

[Mr_mechEngineer]

We could also show that g is continuous by showing that it is derivable; we show that the limit as t goes to a of (g(t)-g(a))/(t-a) exists

[yakovify]

For continuity one can also say that since it's bounded from above by e dominated convergence theorem implies limit of integrals is the integral of the limit which is exactly continuity.

[ingobojak]

The integral evaluates to Sum_( k = 0 to infinity ) [1 / ( k! * (1 + t + k) ) ]. Numerically, t = 0.193155 yields about 1.5.

[giovanicampos]

4:15 I guess you meant g(1) < 3/2 < g(0).

[alexnagle]

I studied math as an undergrad and I never heard of the DI method... it would have made further studies in probability etc. SO MUCH EASIER

[abebuckingham]

At 4:21 you should have g(1) < 3/2 < g(0) since g(1) = 1 and g(0) = e-1 rather than g(0) < 3/2 < g(1).

[dmitrystarostin]

For some reason, this was one of the best points of calculus, probably because it best illustrates continuity for integrals.

[Pichael]

this is a really nice watch. lovely video

[pbj]

2:49 We can do better and prove this fact rigorously by using the power series definition of e^x(=Σ{x^n}{n!})

Putting x=1, we get-

e=Σ{1^n}{n!}=Σ{1}{n!}, n=0 to ∞.

By expanding the first three terms to get 2.5, we find that-

e=1/0! + 1/1! + 1/2! + Terms

Since each subsequent term is positive, their sum must also be +ve and so we can call the sum ε which is >0.

=> e = 2.5+ε >2.5
=> e>2.5
=> e-1>1.5. QED