I envy your students. You are an amazing teacher.

@blackpenredpen

6 жыл бұрын

Aaron Hollander thank you!

I really love your enthusiasm. Thank you for your videos and the magnificent lessons!

@blackpenredpen

7 жыл бұрын

Marian P. Gajda thank you for your nice comment.

I thought I recognized the function. After you got the answer, I noticed this is what we called the "Pert" function for continuously compounding yearly interest. V = P(1 + R/X)^(T*X) P=initial principal R=interest rate T=number of years the money is in the bank X=the number of times a year you compound the interest (1 for yearly, 12 for monthly, 365 for daily, etc) Compounded continuously, X=infinity, so V = P*e^(R*T)

@GeekProdigyGuy

Жыл бұрын

Yup! e is just the interest rate of continuous compounding over 1 unit of time (let's say year), assuming non-compounded 100% interest over 1yr. If you increase the yrs from 1 to b, you get e^b of course, since you just grow multiplicatively by e each yr. And if you increase the rate from 1 to a, you again get the same effect, because you've reduced the time it takes to get to 100% interest w/o compounding to 1/a yrs. Meaning 1 yr at rate a is identical to a yrs at rate 1, so e^a. Combining the two gets e^ab. (You can check that, by symmetry, the same effect happens when a

I'm currently taking Precalculus in High School, but the way you explained it somehow makes me understand it! The videos you've been making are awesome, and I hope you continue showing us things as cool as this!

You explain these things in such a clear and concise way that me being a high school student, I still get some knowledge out of these videos. Hope I have some professors as good as this next year

@blackpenredpen

7 жыл бұрын

Thank you. I hope you the best too. Are you in 11th or 12th grade?

@uchihamadara6024

7 жыл бұрын

blackpenredpen 12th grade, currently learning about optimization in calculus. It's my favourite subject:)

@blackpenredpen

7 жыл бұрын

Uchiha Madara I see! That's great!! I like optimizations too and should definitely make videos on then soon!

Hi, sir! Long time fan here. I watch all your videos for fun and for study. Your calculus videos are awesome and very entertaining. Got very excited when you're doing the infinite series, one of my fav topics in calculus II. Can I make suggestions for future topics for your videos? Can you make videos about the ever interesting stuff about the Bernoulli numbers, gamma function, Zeta functions, q series, Ramanujan Summation? Any stuff related to Real, Complex and Analytical Number Theory with be great for me. Thanks and keep the genius coming!

@blackpenredpen

7 жыл бұрын

Joshua Garcia hi josh, thank you so much for your nice comment. I may not be able to do them anytime soon tho since I have a lot of topics that I want to cover for my students first. But whenever I have time, I will definitely squeeze in some videos about random math problems for fun!

@leif1075

3 жыл бұрын

@@blackpenredpen I took the natural log but then didnt think of lhopitals rule..isnt there ankther way to do it without lhopitals? I really hope you can please respond.

I love the informal pet names that teachers give to certain uhh facts. It's fun and it actually helps students remember important concepts! For instance, my calculus teacher called the chain rule "THE MAGIC". She always said, REMEMBER THE MAGIC.

@blackpenredpen

2 жыл бұрын

Yea, I find it super helpful too!!

I wish you had been my math teacher. Great work. I enjoy your videos even though I haven't had a math class in many decades.

Man, I really wish I'd seen your videos during A levels. You are so fun in your explenations. Now I'm studying mathematics at uni and while your vids are great I don't get the same satisfaction due to the fact that I can tell the answers myself quite easily. :(

I think I have a simpler way of doing this.... Since (1+x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3!........ putting x= a/x and n=bx we get 1 + ba/1! + (ba)^2/2! - b(a)^2/x+ ... and now all the terms with x in the denominator will become 0 as x->infinity so we are left with 1 + ab/1! + (ab)^2/2! + (ab)^3/3!.... this is just like the wxpression for e^x e^x = 1 + x/1! +x^2/2!....... therefore the answer is (e)^ab

@flazzydirect1854

3 жыл бұрын

Makes sense

@ramprasath8091

3 жыл бұрын

Man, that's brilliant!

@leif1075

3 жыл бұрын

But what if you didn't know or haven't proved that infinite expression for e?

@RSvieo

3 жыл бұрын

@@leif1075 you find it with developemtn in serie : exp verifies the differential equation exp = d/dx (exp) so you search a solution serie of general term (a_n x^n) and you find exp(X) = Σ x^n/n! So e = Σ 1/n!

Yes, that's very fine. Just keep posting em and me and my mates'll watch them. We're very entertained. Thanks so much for these videos. You're way fun and captivating than my professors!

you and my cal 2 professor are making cal 2 very easy!! You guys are the best:)

Thank you so much. Your video's concept is crystal clear to me.

we can actually do this without l'hopitals rule by the way your way of explaining is so good!

im so grateful for you!!

You are an amazing teacher. Thank you

@blackpenredpen

7 жыл бұрын

NToB36 thank you for the nice comment! You are amazing too!

or we can say that (1+a/x)^bx = e^(bx*ln(1+a/x)) so when x approaches infinity a/x approaches zero and we can use ln(1+x) = x+ o(x) so L = e^(b*x*a/x) = e^ab !

@uxxlabrute

7 жыл бұрын

I really love your videos btw

@blackpenredpen

7 жыл бұрын

Imaspammedboy thank you

You never fail to impress me!

@blackpenredpen

7 жыл бұрын

Thanks!

love how you call it "The Fact"! 😂

@blackpenredpen

7 жыл бұрын

Ahmed Anwari lol! Thank you. My students love it too! It works

it's known that lim(x-->inf) (1+t/x)^x = e^t now if b>0 just replace by (1+(ab)/(xb)) and replace x by bx, you get e^(ab) I if b

... The " Special " Fact (a = 1 and b = 1) --> lim(x-->inf)(1 + 1/x)^x = e lim(x-->0)(1 + x)^1/x = e ... (lol) ... thank you for a great and very clearly understood presentation, Jan-W

"The fact" is actually a very natural property :D

I enjoy your videos but I'm slightly worried by this one as it could mislead people. You haven't shown the limit is e^(ab). Rather, you have shown that if there is a limit then it must be e^(ab). I would have liked you to have mentioned why the limit must exist.

Hey, we can take the a in bottom and at exponent we do a/a. Like this we have direct e^ab

Salute to you Sir! I'm a huge fan

The result still holds if we replace a & b with sequences that converge to a & b, respectively. This is often encountered in statistics. See the following video. kzread.info/dash/bejne/nWV2yrqtaafNcZc.html

Can't you do it much simpler if you know what e is? what you know: lim x->∞ (1+a/x)^x = e^a and g^pq = (g^p)^q so... lim x->∞ (1+a/x)^bx = lim x->∞ ((1+a/x)^x)^b = (e^a)^b = e^ab

@blackpenredpen

7 жыл бұрын

mrBorkD true. But as I said "do it from scratch" without even knowing "your fact"

@BigDBrian

7 жыл бұрын

I'm just wondering: who would encounter this problem before learning what e is I mean, you use Ln, and limits, even the property that Ln is continuous, and l'hopital's rule, so they must have some idea.

@blackpenredpen

7 жыл бұрын

mrBorkD true. But I know my students. This approach is more suitable for most of them

@BigDBrian

7 жыл бұрын

OK. well I can't argue with you there

@TheYoshi463

7 жыл бұрын

Exactly how I did this :D I mean that way it can be proven with school mathematics (Binomial theorem and power rules, that's all) and done.

You could also have evaluated that limit using a series expansion: ln(1+a/x) = a/x to first order, which gives you the answer in a couple more lines.

I just found this while I was looking for the lowest point ＝1/e for y＝x to the xth power after watching your Chinese video regarding three '0 to the 0th' limit. Looking at the cover I immediately recalled that e to the ath power＝the limit in this video without the 'b', call M. since b is nothing about x but a constant we can take it from the L, the original limit. The limit is now M to the bth power, i.e. (e to the ath power) to the bth power＝e to the abth power.

This Fact, was indeed very well done

Thanks for the great explanations!!! BTW can you also make videos for improper integrals for finding the values of p and q in the condition of convergence?

@blackpenredpen

7 жыл бұрын

Calisthenics Kim I have those already. kzread.info/dash/bejne/pKuAzdSAotjNf84.html U can also go to www.blackpenredpen.com for more resources

I think I need a channel teaching the first few billion years of maths...

you are the best teacher!

We can generalise the function of 1^infinity as Lin X-->inf f(x)^g(x) = e^[{f(x)-1}g(x)]

Here’s a faster way: lim((1+a/x)^(bx)) = lim((1+a/x)^x)^b = (lim((1+a/x)^x))^b (1+a/x)^x = e^a when x appr infinity = (e^a)^b = e^(ab) //

If you sub in a and b to be 1 then you get the equation used to find e (1+(1/n)^n which makes sense coz it’s a^(1•1) which is e.

Can't we also substitute the variables to get the( definition of e) ^ab

The cool accent, mic and apparent joy from calculus make this video. As all others from this guy. Goddamn do I subscribe!

@blackpenredpen

6 жыл бұрын

LOL! Thanks!

Setting a, = b = 1 is a nice way of getting to compound interest tending to e

@blackpenredpen

7 жыл бұрын

yea

We can use lim x tends to infinity (1+a/x)^bx as e^lim x tends to infinity a/x divide by bx So e^ limit x tends to zero a/x*bx Then a /x multiple by bx and x will be cancel our And it becomes e^ab

Thank you for “The Fact” 👌

THE FACT thanks btw, studying for math quiz tomorrow using this !

... that's THE FACT Jack ... !

@blackpenredpen

7 жыл бұрын

WJL what jack?

this is awesome, thank you so much

Hey ... Can you do some number theory problems ?

Thanks so much...the Fact is so useful for me...

The answer if obvious if you recognize this as the limit of the compound interest formula as the compounding period goes to infinity. P(1+r/n)^(t*n) - > Pe^(rt)

Great video, it was very thorough. Hopefully, this trick comes useful on my calc midterm.

@bird9

3 жыл бұрын

yeah that's what I am telling to myself when I learn a new thing like this one...

Okay, so you used a natural logarithm to get bx out front. What if you used any other logarithm (such as log base 10) to do so? Would that not change the answer? Are you only allowed to make everything in your equation the exponent of the base of the logarithm you want to cancel out if it is a natural logarithm?

I like this channel. It states facts.

and what if instead of 1 we had another constant, such as C? the result would be e^(ab/c), right?

How can we define asymptotes of such Functions??

In my personal mnemonics i'm calling this the 'E-fact'

Does that mean every continuous function can go through lim?

What if you take log base 10 instead of ln? would that change the answer?

Even if I was doing this right I'd feel like I was doing this wrong

Put = ay and the result follows at once. I guess he wanted to show the principle.

Would there be a similar limit for complex z, i.e. for some sequence z_n such that for all n(in Natural numbers), there exists some N such that for all m>N, |z_m|>n, the limit lim(n->∞)(1+a/z_n)^bz_n exists and is equal to e^ab?

It was really fun to watch this video I really enjoy it

Other way to do it is to simply evaluating the limit of e^(bx * ln(1 + a/x)). Doing L'Hôpital (of course doing bx = 1/(1/bx), else it would not make sense) would be essentially the same as you did, just without equalling to anything to get an answer; OR you can use the transformation lim(x->c) ln(f(x)) = lim(x->c) (f(x) - 1) if and only if lim(x->c) f(x) = 1: e^(lim(x->∞) bx (1 - a/x - 1)) = = e^(lim(x->∞) bx * a/x) = = e^(lim(x->∞) ab) = e^ab

Great work. Do you have also a simple proof, that f(x)/g(x) equal to (d/dx f(x))/(d/dx g(x))?

Can anyone correct me if I'm wrong, but this formula should still work even if the denominator x, has a constant being added or subtracted, for example x + 2 or x - 5, cause at the end, it will eventually be zero and you will be left with (ab)/(1 + 0), which is just ab

Thnku very much sir❤🙏. It has helped a lot

i don't speak english but i love this guy, i understand better than the videos in spanish

@blackpenredpen

6 жыл бұрын

Thank you!!!!!

How do we know L is positive in order to use ln in it?

Shouldn't the differentiation of [ b * ln ( 1 + a/x) ] be a product rule?

And you can take the ln because you can say that there ist an x0 so that 1+a/x will be positive for all x greater than x0?

I feel like doing the derivative of the numerator and demonstrator doesn't guarantee continuity unless the function in the numerator and denominator are the same. otherwise the delta of one will grow at a different rate and the fraction will no longer represent the same ratio. Same reason why 1/1 = 1^4/1^4 but 4/1 =/= 4^3/1^3 you can't just apply operations willy nilly on numerator and denominator.

Thank you very, very, much.

Why don't you utilise the definition of e.... lim x->inf (1+1/x)^x =e lim x->inf (1+a/x)^bx =lim x->inf [1+1/(x/a)]^[(x/a)(ba)] ={lim x->inf [1+1/(x/a)]^(x/a)}^ba =e^ab which is much easier....

@blackpenredpen

7 жыл бұрын

The reason is I can show this video to my students if they have to do something like, lim as x goes to inf, (1+1/x^2)^x This limit is 1. www.wolframalpha.com/input/?i=limit+as+x+goes+to+inf,+(1%2B1%2Fx%5E2)%5Ex

What if instead of taking ln on both sides you take for example base 10 log or different base log?

@viktor5813

2 жыл бұрын

Then when doing L' Hopital's rule you will need to add ln(10) because of its derivative and it will get complicated

you're just A-OK! Thanx a lot from Paris!!

how about if it is a number other than 1 e.g lim x-> 0 ( 2 + a/x) inside the limit (i presume the answer doesn’t involve e any more? maybe with a natural log of an integer?)

You are great sir Please can you make Indian language videos maths

can you clear my doubt, that 1 to the power anything is 1 because if you keep multiplying 1 to itself its not gonna change. so why 1 to the power infinity is not one?

I want to be your student

The limit as x goes to ∞ of (1+i/x)^(ix) = 1/e by the Fact

It is possible to solve this w/o using l'hopital right? can you please answer me how you solve this NOT using l'hopital? Thanks!

Well explained!

excellent!!! thank you

He is my legitimate god

Thanks! Really good explanation!

@blackpenredpen

7 жыл бұрын

weerman44 you're welcome

How can we know if the limit exists in the first place, cos your derivation assumed it's existence, which happens to be e. Thank you!

why could you take the derivative of the numerator and denominator? do you have a video explaining this (or another good source)?

@tommyliu7020

6 ай бұрын

Look up L’Hopital’s rule

I love your videos What is your name?

You know, I think I see why you like the fact so much...

I still don't understand why ln(lim(f(x))) = lim(ln(f(x))). You said that ln was a continuous function, which I understand, but I don't understand how that yields the swap correct algebraically.

we haven't studied "l'hopital's theorem" so here is how I would do it : I would say let t=1+a/x so then we get lim t->1 ab ln(t)/t-1=ab

amazing

Show that limit x approaches to + infinity then prove ln(1+a^x)/x=1

If I used log base 10 instead of ln I would have a different answer. Why does it have to be ln

just wondering if I replaced the natural log with other logs wouldnt that change 'The Fact'? like if I use lg it would end up 10^ab, no?

@blackpenredpen

7 жыл бұрын

No, the answer will still be the same.

@rodllewellyn

6 жыл бұрын

As I mentioned above: You could use, say, base 10 log or any other base. But, when you use L'Hopital's rule and take the derivative of log(1+a/x), guess what: a ln() will sneak in there haha.

I call this 'The Magical Limit Escape' When all the checks, La Hospital Rule or any other rule don't lead to me any conclusion, this formula (according to you) works as a magic and leads to the exit way!!!

heavy duty !

If I chose to take the log of both sides, wouldn't I get 10^ab as my final answer?

@SuperMerlin100

7 жыл бұрын

No because when you used the LH rule the derivative of logx isn't 1/x. /1(xln(base))

@Uncle_Yam

7 жыл бұрын

Oh okay, thanks!

you took the derivative of the right side. souldnt you have to take the derivative of the left side (ln(L)) also??

@sebastiansimon7557

5 жыл бұрын

He didn’t take the derivative on the right side, he used l'Hospital on the right side which, if applicable, doesn’t change the value. The derivative of f(x)/g(x) would use the quotient rule which isn’t f'(x)/g'(x).

It's just..strange, you used log base e, to prove the definition of the number e, is it okey?

You don't have to use a natural log for the calculation though, right? Would you get to the same result if you used a normal logarithm?

@rodllewellyn

6 жыл бұрын

You could use, say, base 10 log or any other base. But, when you use L'Hopital's rule and take the derivative of log(1+a/x), guess what: a ln() will sneak in there haha.

@Filip6754

6 жыл бұрын

Roderick Llewellyn Fascinating

I would not put the limit L because we do not know if the limit exists a priori..maybe this techniques gives a correct result in this case ,but in general it is not correct(for pathogenic functions)..If f(x) is the function mentioned in the video i would write: f(x)=e^ln(f(x)) and the compute the limit of ln(f(x))

Could someone explain why f(lim g(x)) only equals lim(f(g(x)) when f(x) is continuous? in this case, f(x) = lnx.