Math Slander

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Math Slander

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"The proof is left as an exercise for the reader" goes so hard

igormattos

all terms cancelling to make 1=1 is my worst nightmare, shit haunts me

Readraid_

As someone who doesn't like maths, i didn't expect the slander to be much easier to understand.

rage

As someone who actually missed the passing mark by one point because forgot to put +C I felt the pain

justpassingby

Sad Spiderman relates to all students leaving after a math test

Caleb-DH

This week i was calculating an equation on a test, made 6 mistakes in the process but got the right answer, how? The mistakes cancelled themselves out.

petermikus

As an engineer I felt those two clips starting at 1:52.
Sometimes I just get curious on how something was/can be proven but it is only worth 10min of my time.
But the math books will be like: "Proof is left as an exercise 🗿" It is so frustrating especially when you didnt or dont study maths.

staticnullhazard

2:20 If anyone's curious, you use logarithmic differentiation (where you take the natural logarithm of both sides before differentiating). This tool is used whenever you need to take the derivative of anything complicated

gabedarrett

2:23 this is always true, because the people who aren't good at math have trash grades, and the people who are good at math judge themselves too harshly and think that a 99.9% is failing lol

DoomRutabaga

How is trigonometry and calculus going to help me pay my rent?
Teachers: yes

Name-xdhv

0:23 literally the truest thing this year

chrisbl

As someone whose maths abilities is that equivalent of a chimp, I struggle to grasp the concept of the meme/slander.

birinderwarraich

For 0:55, I am a madlad who actually spent 30 mins to work out the answer.

For simplicity, just consider when x > -1 so that we won't get complex number. The solution is given by

/ (x²+1)^x

Let me explain the steps.

First of all, note that the n/(n!)^(1/n) term can be completely disregarded since on its own, it's limit is 1. The proof is left as an exercise to the reader. Also, let's ignore the x in the exponent for now since it can be easily added back to our answer later

Then we are left with is

f(x) = Π [(x+n/k)/(x²+(n/k)²)]^(1/n)

(Π is capital Pi) where k = 1, 2, ..., n as n → ∞. This strongly encourages to take log on both sides, ie we define

g(x) := ln[f(x)]

= Σ{ ln[(x+n/k)/(x²+(n/k)²)] / n}

This is very similar to a Riemann Sum and hence we yield

g(x) = ∫ ln[(x+1/t)/(x²+1/t²)] dt

from 0 to 1. This motivates us to use the substitution u = 1/t, which gets us

g(x) = ∫ ln[(x+u)/(x²+u²)] / -u² du

from ∞ to 1, which is

∫ ln[(x+u)/(x²+u²)] / u² du

from 1 to ∞. With the natural log inside the integrand, we hope to use integration by parts to remove it, hence

g(x) = ∫ ln[(x+u)/(x²+u²)] d(-1/u)

= [ln[(x+u)/(x²+u²)]*(-1/u)

+ ∫ 1/u d(ln[(x+u)/(x²+u²)])

Evaluating the limit of the first part, we get ln[(x+1)/(x²+1)]. For the latter, we calculate the differentials with the product-to-sum rule of ln to get

∫ 1/u d(ln[(x+u)/(x²+u²)])

= ∫ 1/u*[1/(x+u) - 2u/(x²+u²)] du

= ∫ 1/u(x+u) du - 2 ∫ 1/(x²+u²) du

The former can be easily dealt with partial fraction to get

∫ [1/(u+x) - 1/u] / -x du

= [ln|1+x/u|] / -x

Taking limit we get ln|1+x|/x. The latter can be dealt by trigonometric substitution, ie

- 2 ∫ 1/(x²+u²) du

= [-2/x*arctan(u/x)]

Taking the limit as u → ∞, we get arctan(∞), which should be π/2. but we also see 2 x whose signs cancel out, thus we get - |π/x| + 2/x*arctan(1/x). Finally, we get

g(x) = ln[(x+1)/(x²+1)] + ln|x+1| /x -
|π/x| + 2/x*arctan(1/x)

By using f(x) = exp(g(x)) and finally adding the x back into it, we get out final solution as shown at top.

(It is 3:30 am in my city and wtf was I doing?)

aidancheung