How Do You Actually Compute Logarithms?

[@BriTheMathGuy]

Become a Math Master With My Intro To Proofs Course! (FREE ON KZbin) kzbin.info/www/bejne/aZTdmJl-irGNedU

[@MathTidbits]

another homemade approximation formula for Ln ( a + b ) ; where a is an integer ; 1 < a < 10 and b is a fraction in decimal form . thus for 1.5 ; a= 1 , b = .5 2.3 Ln ( a + b ) ~ Ln ( a ) + ___________________________ 1.07 + 2.3 ( a / b ) Ln ( 1 + .5 ) ~ Ln ( 1 ) + 2.3 / [ 1.07 + 2.3 ( 1 / ,5 ) ] Ln ( 1 + .5 ) ~ 0 + 2.3 / [ 1.07 + 2.3 ( 2 ) ] Ln ( 1 + .5 ) ~ 0 + 2.3 / 5.67 Ln ( 1 + .5 ) ~ 0.4056..... Ln 1.5 = 0.405465... ( true value ) note; it is just a rough approximation,but I think it is more useful in the Log (a + b ) form.

[@mikecaetano]

I'm old enough to remember the time before hand held calculators. We used lookup tables for logarithms. Slide rules were before my time. ln(1.5) = ln(3)-ln(2)

[@ayuballena8217]

yes

[@Ninja20704]

What I always like to say is that these are the same algorithms that calculators use to give you your answers. Not just logs, but also radicals, trig, exponentials etc (Ok maybe not the same algorithm, but some algorithm) Because remember that they too also don’t know how to do anything more than basic arithmetic(addition, subtraction, multiplication, division). The only difference is that they can do it many times faster.

[@robread-jones3698]

Simpson's Rule eh? I'll remember that rule.

[@Jovian_Man]

*BREAKING ON PROGRESS*

[@alixx_legenddark_xx2819]

A Simpson comment eh? I’ll remember that comment.

[@dudz1978]

Easy method to calculate decimal log of x (1

[@dudz1978]

Example: log(2) =(1/10)log(2^10)=(1/10)log(1024)=(1/10)(3+log(1.024))=0.3+(1/(10^2))log(1.024^10)≈0.3+(1/(10^2))log(1.26765)=0.3+(1/(10^3))log(1.26765^10)≈0.3+(1/(10^3))log(10.71504)=0.3+(1/(10^3))(1+log(1.071504))=0.301+(1/(10^3))log(1.071504), etc.

[@angeldude101]

This should work for any Natural number base. Just replace "10" with whatever "10" would mean in the desired base. This requires being able to convert between bases, but that's a fairly simple algorithm, especially for integers.

[@powerdriller4124]

Not as good for doing it on paper, but with an old 12digit calculator having square root function, the following method works charmy smooth for 1

[@lizzybach4254]

I'm in grade 9 and under the SEBA board, you're required to take an elective subject. Mine is "Advanced Mathematics" and in chapter 3, we're actually introduced to logarithms. A great bulk of the chapter is actually about finding logarithms while being given some or no values. Edit: I need to mention that obviously we don't get calculus and other complex stuff in grade 9.

[@ericbischoff9444]

A third solution with basic arithmetics: ln 1.5 = x means e ^ x = 1.5 Let's try the square root (x = 0.5) 1.6 x 1.6 = 2.56 and 1.7 x 1.7 = 2.89. That means that sqrt(e) must be between 1.6 and 1.7 in order to reach 2.71. so x = 0.5 is a bit too big because it gives, say, 1.65 instead of 1.5. Let's try 0.4 then. Let's compute 1.5 ^ (1 / 0.4) and see if we are close to e. 0.4 is 4 / 10 so its inverse is 5 / 2. So 1.5 ^ (5 / 2) is sqrt(1.5 ^ 5). 1.5 is 3 / 2, raised to power 5 is (3 ^ 5) / (2 ^ 5) = 243 / 32. Just notice that 243 / 32 is almost 8 and sqrt(8) = 2 x sqrt(2) = 2 x 1.4 = 2.8. We indeed get close to e. Alternatively, computing more accurately, 243/32 = 7.6 and sqrt(7.6) = 2.7. OK, very artisanal and approximate, and I was lucky to pick the right numbers... But it works - and you don't even need a pen and paper.

[@maniamhungry4896]

I have been looking for this solution. Thanks man. You saved my life time.

[@symmetricfivefold]

5:06 for some reasons, eventhough i know a little about integral, this looks like enchanting table language to me

[@monsieurLDN]

That's the result of it being overwhelming for you it will become easier eventually

[@atlas_19]

I derived a pade approximation, which is like a convergent taylor series that stays true to the approximated function for a pretty long time, dor natural log after I had learned about them, so here it is: c = x² + 4ax + a², lnx ≈ lna + 1 - 6a²/c where a is the point you're approximating around. If we choose a = 70 the error reaches 0.05 just before x=104 and 0.1 around x=121. For a= 100 error reaches 0.05 around x=147 and 0.1 around x=173. It diverges quite faster for values of x lower than the approximation point and when the approximation point is closer to 0 but I'd say it's good enough for use. I'll post a desmos link in the replies.

[@farklegriffen2624]

For the Taylor series, just replace n with -n, and it converges for |x| > 1

[@angeldude101]

I don't remember ever learning Simpson's rule, but it looks similar to breaking the integral into a discrete sum. What i was _expecting_ to see was how to calculate an integer approximation for the logarithm of an Natural number base. To do that, convert the number to the base in question (which can be done with division and remainder, or even just subtraction and comparison), and then simply count the number of digits to the left of the radix point, or the number of zeros to the right of the radix point before the first non-zero digit. Simple counting and base conversation. No need for complex summations. The trade-off is that it's only accurate to the integer (and may be off by one with just what I described) rather than getting more and more precise. After having the integer value, it's sometimes possible to guestimate roughly where the fractional part is between 0 and 1 based on the remaining digits.

[@herbcruz4697]

I learned Simpson's Rule back in Calc II. It's a useful technique for calculating integrals numerically (Of course, the number of subintervals needs to be even, in order to use Simpson's Rule), and uses parabolas to calculate the area under the curve, as opposed to rectangles.

[@TejasIsAmazing]

You "could" also approximate it with newtons method. Lets say we wanted to find ln(c) You could write it as x = ln(c) => e^x = c => e^x - c =0 Now, its derivative is also e^x. So using newtons method, we have a recursive algorithm to approximate x x_(n+1) = x_n - (e^(x_n) - c)/(e^x_n) => x_(n+1) = x_n - 1 + c/(e^x_n) And you can approximate e^x with its taylor series, which does converge for larger values of x, unlike logs. This method is probably much slower for less precision, but since its quadratic convergence, if you want extremely high precision for large numbers for whatever reason, you can approximate it with this. Though at that point you should probably just use a calculator.

[@thataialaperrera810]

Brooo I’ve been looking everywhere for a video on this thank you bro 🙏 😭 ❤

[@MathTidbits]

here is another home made approximate formula for Ln ( 1+x ); where x is 0 < x < 1 Ln ( 1 + x ) ~ (2.3) / [ 1.097 - .085x + ( 2.31/x ) ] ; for Ln ( 1.5 ) = Ln ( 1 + .5 ) Ln( 1 + .5 ) ~ 2.3 / [ 1.097 - .085*.5 + 2.31/.5 ] Ln ( 1 + .5 ) ~ 2.3 / ( 1.097 - .0425 + 4.62 ) Ln ( 1 + .5 ) ~ 2.3 / 5.6745 Ln ( 1 + .5 ) ~ .4053...... Ln ( 1 + .5 ) = .405465.......(true value) For the Log ( 1 + x ) with base 10 ; replace numerator 2.3 with 1.

[@ChadTanker]

I just did the most natural thing to do and keep the log ln(1,5) = ln (3/2) = ln(3) - ln(2)

[@2K.7b]

Great job. Would you mind telling me what app do you use to animate the mathematical expressions and terms?

[@rakib_hoque]

ln(x)=lim(((x^h)-1)/h) , h->0 The limit approximates ln(x) if you put in small like really small values of h I think plugging in 0.0001 for h is probably going a good approximation

[@baadibaadi-k9n]

can you tell me please what logicial did you use to make this fantastic video thanks in advance

[@RunTheProgram]

Great video!

[@keitha199]

Wait, how does the x_0 becomes 1 and x_1 becomes 1.25?? Help i really dont know how does this sub interval thingy works

[@Ninja20704]

You are dividing the region [1,2] into 4 equal regions since n=4. So the first interval is [1,1.25] and the next is [1.25,1.5] and so on.

[@keitha199]

​@@Ninja20704I see it now!, Tysm for teaching me how does the sub intervals works!

[@angeldude101]

​​@@Ninja20704For those of us who either didn't learn Simpson's rule or had forgotten it, this would've been _really_ useful to have in the video itself. Nowhere was this ever explained. Edit: Where did the coefficients of 1, 4, 2, ..., 4, 1 come from? They're all powers of 2, and it _seems_ to be symmetrical, but what's the actual pattern? How do you get the coefficients for more than 5 terms?

[@Ninja20704]

@@angeldude101 you can literally find explanations and/or derivations for simpson’s rule anywhere else online. He did not go through the derivation because that isnt the point of the video. Regarding the coeffecients, the first and last are both 1. For the all the ones in between it alternates 4,2,4,2,…,4 starting from the second one. And one condition is that the n you choose must be an even number.

[@michaelbaum6796]

Thanks a lot for this great explanation 👍

[@Ноунеймбезгалочки-м7ч]

2:23 aren't these supposed to be factorials? you're divinding by 1 2 3 4 but it's 1 2 6 24 no?

[@dominicellis1867]

I’m trying to find the roots of tanx - x, is there a way to use eulers method to derive an ODE and use inversion to find the roots. Newtons method involves polynomials of contangents and the power series is useless because I don’t know where to center it.

[@Neilcourtwalker]

You gotta love the Simpsons and their math. ;-)

[@brahmanandamperuri5721]

Can anyone tell that how f(x1)..f(x4) had taken

[@05degrees]

How does Simpson’s rule in this case compare to a naive rule of just summing (1/1 + 1/(1 + (x−1)/n) + 1/(1 + 2(x−1)/n) + ... + 1/x) (x − 1) / n? (Because it could end up that for this specific case of ln x, not much accuracy might be lost.) Hm I need to check in in Python myself probably. Because the naive rule looks a bit more handy for human calculations.

[@Ninja20704]

Is what your talking about the riemann sum i.e. approximating area just using rectangles? If so then it’s not the best because you need quit a bit more terms to get the same amount of accuracy. What I hear is that simpson’s rule is one of the “best” approximation method for integrals in terms of how fast it converges to the actual area.

[@05degrees]

@@Ninja20704 Yeah, after looking at formulas, I thought that more terms isn’t really any simpler than changing the coefficients!

[@annszymanski2369]

Great video 👍

[@CatInABaseballCap]

but ln(1.6) is even better

[@andunyaa]

Amazing Video

[@Ensign_Cthulhu]

First you calculate ln(10) to some stupid level of precision. Then you calculate the natural logarithms of 2, 3 and 7. From these, calculate the base 10 logs (ln(a)/ln(10) = log(a)). It doesn't take that many starting values on a base 10 log table to fill in quite a few gaps, and the number of horrendous long divisions or multiplications you have to do is minimized. I wonder if having this available might not have saved Henry Briggs a lot of work. Alas, a cursory search indicates that Simpson's Rule post-dates his death.

[@chrisMUC69]

x(0) = a, x(1) = x(0) +∆x, ..., x(n) = x(n-1) + ∆x or x(n)= x(0) +n * ∆x or x(n) = b

[@JonnyMath]

Nice video!!! I also have a maths channel but I'm not as talented as you about maths and youtube!!! Don't give up your videos are amazing!!!🎉🎉

[@vikramadityakodavalla3795]

give up on what, bro u might think its encouraging but sometimes it might convey a different message like "ppl dont like ur videos much but dont give up" im not saying thats ur intention but if u frame it differently it could avoid such problematic threats

[@JonnyMath]

@vikramadityakodavalla3795 Sorry... I simply love this channel and it inspires me to grab my maths textbook and learn maths!!!

[@JonnyMath]

I always write don't give up because we aren't getting a lot of views as maths channels recently, for example, flammable maths and bprp weren't getting a lot of views during summer holidays, so my "don't give up" simply means be motivated and believe in what you're doing because people need your videos and they find them interesting and useful!!!

[@robertveith6383]

​@@vikramadityakodavalla3795 -- Please write in English sentences. Do not use texting.

[@test_dithered9860]

Impressive

[@dakshhcr2574]

Just get me a log table and I'll do it, lmao

[@tanavgowda9512]

hi

[@tamilselvanrascal5956]

🔥🔥🔥

[@ThanhNhan_GiaSu234]

Hi❤❤❤

[@navsha2]

btw log base 2 and 3 is actually 0.03

[@yasuki416]

Thanks alot, i was facing hard time with logarithm in my classes just the other day lmfao

[@thedogfather1204]

🙏

[@theloganator13]

"We can easily compute things like 2^3 by hand, but when it comes to things like log base 2, we reach for a calculator" That's because you're cherry-picking values that are easy and hard to compute. log_2(1) = 0, log_2(2) = 1, log_2(4) = 2, log_2(8) = 3. Not so hard. Can you so easily compute things like 2^2.7? Or 2^3.2? No, because those are hard for the same reason that log_2(3) is hard. Your video does not actually show how to compute log_2(3) by the way. I wonder if it's because both of your methods would require you to know ln(2) a priori 🤔

[@gabrielleyba2842]

log2(3) is not 8

[@robertveith6383]

Make it clear what the base is and what the argument is: log_2(3), log[2](3), or something along those lines.

[@foxypiratecove37350]

No, but log_2(8) is 3 because 2^3 = 8

[@NXT_LVL_DVL]

You should have chosen bigger primes. Almost every math guy knows log values of 1,2,3,5,7

[@robertveith6383]

False. What you wrote is not true.

[@cringotopia8850]

First here 🌚

[@robertveith6383]

Use ln(1 + x) ~ x - (x^2)/2. Write ln(1.5) = 128*ln[1.5^(1/128)]. Using the square root button seven times: About 128*ln(1.003172719) ~ 128[0.003172719 - (0.003172719)^2/2] ~ 0.405463... Compare this to ln(1.5) = 0.405465...