Defining the Natural Logarithm as an Integral?!?!?

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  • čas přidán 17. 07. 2024
  • We typically define ln(x), the natural logarithm of x, by first defining the exponential function of x, noting that this is a 1:1 function, and then defining ln(x) as the inverse function to exponential. In this video we go the other way around. We define ln(x) as a particular integral, and then use that to give a new definition of e. This integral definition of logarithm has several advantages such as being able to be approximated nicely and easily proves that the logarithm function is differential. We also recover standard properties such as ln(ab) =ln(a)+ln(b).
    This video is part of my playlist on Calculus II: • Calculus II (Integrati...
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Komentáře • 41

  • @samisiddiqi5411
    @samisiddiqi5411 Před 3 lety +13

    This is truly a superior approach because it means a lot for computation-- It's honestly better than having to deal with series.

  • @AbdArahman
    @AbdArahman Před 3 lety +10

    You are such a classy teacher who i was looking for.
    This was really helpful.
    Thank you sir .

  • @mdhedaetulislam9009
    @mdhedaetulislam9009 Před 4 lety

    really impressed at your presentation and thoughtful endeavour

  • @zuggrr
    @zuggrr Před 2 lety

    This is awesome ! thank you sir.

  • @AG-hl7bg
    @AG-hl7bg Před 4 lety +2

    Are you going to continue the Laplace transform series? Thank you. 😊

  • @pipertripp
    @pipertripp Před 4 lety +3

    This was really cool. Once you know some calculus, it's a really intuitive way to think about the log function.

    • @pipertripp
      @pipertripp Před 4 lety +1

      yeah, and so often logs and exponentials are introduced in a pre-calc courses long before even limits are introduced. I don't even remember defining the exponential function in terms of limits in my calculus courses... but that was 28 years ago so maybe I just forgot.

    • @angelmendez-rivera351
      @angelmendez-rivera351 Před 3 lety +1

      @@pipertripp Agreed, and I completely disagree with introducing the exponential and logarithmic functions in pre-calculus courses, because it just makes them very confusing to understand.

    • @pipertripp
      @pipertripp Před 3 lety

      @@angelmendez-rivera351 I think with more context it could be ok. I remember learning logarithms in high school in the 80’s and looking up values in tables and having no idea what the hell I was doing. There was no backstory or motivation. I had no idea why people invented them.I was just learning the mechanics. It could have been done so much better.

    • @angelmendez-rivera351
      @angelmendez-rivera351 Před 3 lety

      @@pipertripp Indeed.

  • @BoZhaoengineering
    @BoZhaoengineering Před 4 lety +2

    This is not taught in calculus textbook. But exponential and logarithm are the typical function and inverse function and people use the properties to simplify computation.

  • @maxpercer7119
    @maxpercer7119 Před 2 lety +2

    and in general, for any function, for any input there is one and only one output - this property is sometimes called well defined-ed-ness.

  • @mtaur4113
    @mtaur4113 Před 2 lety +1

    Logs and exp are a real chicken-and-egg definitions game. There are at least three starting points, and what's a "definition" and what's a "theorem" depends, but at the end of the day it's all the same. Who's to say which one's the "definition?" I kind of like the power series definition as a starting point for exp, but this requires some limit exchange theorems to really verify anything. Luckily, you don't need exp to prove what you need to know about power series.
    In any case, the road from A to B might be long, but well-defined and non-circular. But the same works from B to A. Or going from A to C, or whatever. Historically, and how they teach it in high school, might be drastically different from the "name that tune" version of the definitions game, where you start with the tidiest definition and try to prove all theorems in the fewest number of steps. "Name that tune" approaches can be counterintuitive for first-time learning, but great for second and third looks.

  • @MrMarkgyuro
    @MrMarkgyuro Před 4 lety +6

    Trefor could you explain me the change of variable steps with u=t/x, i always struggle with this. :-)

    • @MrMarkgyuro
      @MrMarkgyuro Před 4 lety +2

      i figured it out :)

    • @bliezz
      @bliezz Před rokem

      I have the same problem here, could you explain it please? Or what material can i read to understand that?

  • @lucielpires705
    @lucielpires705 Před 2 lety

    Consider the region R delimited by the curve y = C− x^2 and the x axis. Use integration to find the value of C > 0 so that the volume of the solid obtained by rotating R about the x axis is 64√2/15π
    Note: Plane sections known to the x axis are circles. Would you help me?

  • @rickykris8840
    @rickykris8840 Před 4 lety

    wow its so special
    l've never seen it before

  • @raymondreddington6317
    @raymondreddington6317 Před 3 lety +2

    You are great

  • @SHAHHUSSAIN
    @SHAHHUSSAIN Před 4 lety

    Nice content👌

  • @whyyat3470
    @whyyat3470 Před 2 měsíci

    I don't get that substitution at @9:19. I can't seem to get to that du. Maybe it was u= x/t ???

  • @salikmalik7631
    @salikmalik7631 Před 4 lety +1

    Find your channel today. really Awesome content.
    Do'nt you have lectures on statistics ?

  • @trueresources3847
    @trueresources3847 Před 3 lety +1

    at roughly 10:00 where you used u=t/x, why didn't you changed the dt to x*du?
    please tell if I'm missing something

    • @Michallote
      @Michallote Před 2 lety +1

      It is not needed as the variable t matches with the dt and we can apply the previous definition of ln

    • @xtasu
      @xtasu Před 2 lety

      Same here

    • @xtasu
      @xtasu Před 2 lety

      @@Michallote What?

    • @squeezy8414
      @squeezy8414 Před rokem

      The full substitution really should've been t = xu
      - dt/du = x => dt = xdu
      - t = x => u = 1
      - t = xy => u = y
      - The integral becomes the integral from 1 to y of (1/xu) * x * du
      - This simplifies to the integral from 1 to y of 1/u du, which by definition is this ln(y) value.

  • @davethesid8960
    @davethesid8960 Před 3 lety +1

    Woah, what happened at 2:31?

  • @666frank8
    @666frank8 Před 4 lety

    Is this the last video for Calculus II?

  • @Dreamprism
    @Dreamprism Před 3 lety +1

    Ah - of course. Split up the integral like that to prove the ln(xy) = ln(x) + ln(y) rule with this definition of ln.
    I was trying to do something much more complicated, and it just didn't feel right.
    Thanks for the video - even if I may have fast forwarded just to get the the part I needed. ;)

  • @yourfutureself4327
    @yourfutureself4327 Před rokem

    💚

  • @xtasu
    @xtasu Před 2 lety

    9:19 I don't know how you guys in the comments figured it out but I still can't get over that substitution there. Why is it not (1/u)*x*du and just (1/u)*du?

    • @squeezy8414
      @squeezy8414 Před rokem +3

      The full substitution really should've been t = xu
      - dt/du = x => dt = xdu
      - t = x => u = 1
      - t = xy => u = y
      - The integral becomes the integral from 1 to y of (1/xu) * x * du
      - This simplifies to the integral from 1 to y of 1/u du, which by definition is this ln(y) value.
      Hope that helps

  • @dagraves
    @dagraves Před 2 lety +2

    I don't understand how you got 1/t dt = 1/u du

    • @dagraves
      @dagraves Před 2 lety +1

      I figured it out, thanks.

  • @mhsn27mhsn10
    @mhsn27mhsn10 Před 4 lety

    🖒

  • @greger333
    @greger333 Před 4 lety +2

    Try to decrease movement of your hands and arms. It is very distracting in the long run.