The number e explained in depth for (smart) dummies

The number e explained in depth for (smart) dummies

Welcome to another Mathologer video. A while ago I did this video in which I wanted to explain the mysterious identity e to the i pi is equal to minus 1 to Homer Simpson who encountered it in one of the Treehouse of Horror episodes. So my mission was to explain why this identity is true to someone who knows only basic arithmetic: plus, minus, times, divided. Today I’d like to do the same for the other fundamental facts about the number e. These fundamental facts I’m thinking of include this truly amazing identity here that expresses the exponential function as an infinite sum. I’ll first show Homer and you how you can derive this identity from scratch just using basic arithmetic. Then we can set x equal to 1 and use the resulting sum to calculate an arbitrary number of digits of e. Now here the main point is to figure out when you can stop adding the terms of this infinite sum and be absolutely sure that you’ve calculated the first, say 1 million digits of e or however many digits you are after. Then I’ll show you that e is an irrational number, why e cannot be written as a fraction. And for the finale I have to catch Homer on one of those days that he’s super smart, that he knows a little bit of calculus for example. The finale is about showing why the exponential function is its own derivative, some super famous facts about the natural logarithm, the inverse of the exponential function and a couple of other super cool insights to round things off. Now the main point of the video is to show you WHY all these things are true and not just talk about them like it’s usually done on YouTube and to show this to you in as elementary and accessible way as possible. As usual please let me know in the comments how I did in this respect. In the e to the i pi video I first showed Homer that if he invests one dollar at an interest rate of one hundred percent and compounds the interest N times during one year he’ll have that many dollars at the end of the year. For example, if you compound twice you end up with two dollars and twenty-five cents. Now the more often you compound the interest, that is, the larger the N, the more money you’ll have by the end of the year. But this amount does not grow forever. There’s a limiting value and this limiting value is the number e which is approximately 2.718. Then I show Homer that this actually extends in a straightforward way to the exponential function. So plug in any value for x like for example 5, crank up the N and in the limit you get exactly e to the power of 5. Then we just go for it and plug in i pi for x and see what happens when we crank up N. Translating all this into geometry leads to this neat complex number pokeball there. It looks like a pokeball right? The green point shows what complex number we are dealing with for N=6. Then cranking up N you can see the green point tending towards -1 and so e to the i pi is equal to minus 1. Super neat, check it out 🙂 For today I’ll return to this point of the argument here and branch off. Now let’s try something super natural. I think anybody would try this. Let’s expand the right side for a couple of small N and see whether we can see some sort of pattern. So N equals 1, nothing to expand here. N equal to 2, I’m sure everybody watching can do this. N equal to 3 and equal to 4, and so on. A couple of patterns are emerging, right? In terms of green and blue bits it’s clear how this continues. What about those other numbers? Well, there’s also a pattern and I’m sure many of you will be familiar with it from school. These are binomial coefficients and can be written like this. We’ve got our pattern right there. But before we carry on just a reminder that there is a special way to write and refer to the denominators 1, 1 x 2, 1 x 2 x 3, and so on. For example, 1 x 2 x 3 is written three exclamation mark and it’s called 3 factorial. And there we have 1 and 2 factorial. Now that we’ve established a pattern let’s crank up the N and replace 4 by something big, say 1000. All right, so here we go. Now what I want to do is swap things like that. And what I mean by this is I want to swap 1! by 1000, 2! by a thousand squared and 3! by a thousand cubed, and so on. So let’s just do it. Now let’s have a look at the fractions. Okay this guy here is equal to 1. That one is equal to 1, too. What about that one here? Well it’s close to 1 and as we replace 1000 by larger and larger values this fraction will tend to 1. All the other fractions that you see here also go to 1 and so all the numbers here will go to 1. And so in the limit we get this identity here. And if you are a mathematician you can dot all the i’s and cross all the t’s to prove that this amazing identity is really always true no matter what you plug in here for X. If you know more maths you may have seen this identity before but derived in a very different way and in a more general context. In that general context it’s called the Maclaurin series of the exponential function and there it is derived using calculus. Setting X equal to 1 we get this identity here and it allows us to approximate e by adding more and more of those simple terms. Now, as promised, and i hope Hormer is still with us, there’s only basic arithmetic and some glossing over details involved. Before we calculated e by cranking up the N in this expression here. In terms of using this expression to actually calculate a good approximation for e the problem is that every time we up the N we have to start afresh and discard everything that we’ve done so far. More importantly, although we know that we get e in the limit, to start with we have no idea how large an N we have to choose to be able to guarantee that we found e correct to the desired number of digits. On the other hand, the infinite sum is much much more user-friendly in this respect. As we add more and more of those simple terms our approximation gets better and better and it’s also very easy to estimate how well an approximation a chopped-off partial sum is. So here we’ve chopped off the infinite sum at the 8th term. How close does this get us to e? Well, the error or the difference between e and this approximation is just the sum of the remaining terms. Let’s estimate how large this error is. The 8! at the bottom is just 1 x 2 x and so on to 7 which is 7! times 8. 9! is 7! times 8 times 9, and so on. Let’s pull out the 1/7! Now the bracket is still very complicated so let’s do something drastic, let’s replace the 8, 9, 10, etc. in the denominators by 2s. This gives the new simpler expression. Is this new simpler expression greater or smaller than the error? Well we’re going down in all the yellow denominators, so that means that the new expression has to be greater, right? But now it’s really easy to see that in the brackets we have 1/2 plus 1/4 plus 1/8, and so on which you all know is equal to 1. And so to summarize, the error we make by chopping the sum off at the 1/N! th term is equal to 1/N! which is very good because 1/N! gets very small very quickly. So let’s just put down all the numbers in sight here. If we focus on the zeros in the estimate for the error it seems clear that our approximation will be correct in the first four digits, right, and it is. So this means that if we wanted to have some fun calculating the first 1 million digits of e from scratch, to pinpoint how many terms of the infinite sum we’d have to add we simply have to figure out which number we have to substitute 7 by to get a million zeros at the beginning of the estimate at the bottom. Can someone discuss in the comments why, just going with what I’ve shown you so far, one might want to aim for a million and one zeros instead of just a million. Anyway, in this way you can figure out that you have to add a little more than the first 205,000 terms of the infinite sum to be able to guarantee a million digits of e, very neat, right? And maybe surprisingly the sort of thinking that went into all this is also very applicable. In fact, you use infinite sums and estimations like this all the time to come up with approximations of complicated numbers that are accurate enough for practical purposes. It also turns out, and this really is very surprising I think, that this estimate for the error also gives a straightforward way to prove that e is an irrational number, that no fraction is equal to e. Are you ready for some real magic? Okay, so let’s summarize what we’ve done so far like this. On the left side we have the difference between the true value and the approximation, that’s the error, right? And we’ve seen that this error is less than 1/7! Now let’s warm up by using this inequality to show that one particular fraction, 19/7 is not equal to e. Why 19/7 ? Well, it does not really matter what fraction we start with. The only reason for us using 19/7 is that it has a 7 in the denominator which will mesh in nicely with the other 7s that are already floating around here. Can e be equal to 19/7 ? Well, let’s assume it is. Then we’ve got this inequality up there. 19/7 minus the junk in the brackets is less than 1/7! Okay, now all the denominators on the left divide 7! , right? Just think about it for a moment. Yes they divide 7! which means that we can combine the left side into a fraction with denominator 7! where the ? stands for 1 or 2 or 3 or some other positive integer. So IF 19/7 was really equal to e, THEN we just showed this fraction, the error would be 1/7! or larger. But, and this is really the punchline, we already know that the error is definitely smaller than 1/7! . So our assumption that 19/7 is equal to e implies a statement that is obviously false. So this means that our assumption, that 19/7 is equal to e must have been false to start with. Tada! Okay you probably didn’t see that coming, right? To show that e is not equal to any fraction a/b we simply have to adjust our argument like this: first replace 19/7 by a/b. Then, instead of summing up to 1/7! we sum up to 1/b! . The estimate then also changes to 1/b! and this statement down here is still impossible which implies that no fraction can be equal to e or, in other words, that e is irrational. Now you may have to watch this part of the video again to really get what is going on here. What makes this proof work is the fact that the close approximation to here on the left is super close in the sense that it can be written as a fraction that differs from e in less than 1 over its denominator. Similar super close approximations to pi and other irrational numbers also play a crucial role in proving the irrationality and transcendence of these numbers. Actually a video on transcendental numbers is next on my to-do list, so stay tuned. The Simpsons singing nonsensical stuff… (announcer) Tonight’s Simpsons episode was brought to you by the symbol Umlaut and the number e, not the letter e but the number whose exponential function is the derivative of itself. (Mathologer) Hmm, the exponential function is the derivative of itself. Before I deal with this congratulation to Homer and you for making it this far. For this last part of the video I’ll assume that you do know a little bit of calculus. In particular I’ll assume that you know what the derivative of a function is. If you need an introduction or a refresher check out this video up there and then return here or just sit back and admire the mathematical magic doing its thing. Okay we want to convince ourselves that the exponential function is the derivative of itself. Because of this identity the derivative of the exponential function is just the derivative of 1 plus the derivative of this term plus the derivative of that term, and so on so. What’s the derivative of x cubed divided by 3! for example? Let’s see: 3! is 1 times 2 that’s 2! times 3. The derivative of x cubed is 3 x squared. The 3s cancel and of course what remains is equal to this term here. So the derivative of x cubed divided by 3! is simply the previous term and you can check that the derivative of that previous term is the term before that. And the derivative of that term is 1 and the derivative of 1 is 0, and so on. And so you can see that the derivative of all the junk on the right side is again the right side which means the exponential function is its own derivative. Fantastic, right? So, if you made it up to here there is no reason to stop. Let’s use this insight for something fun. The natural logarithm log X is the inverse of the exponential function. This means that e to the log X is equal to X. Let’s find the derivative on both sides. The derivative of X is 1 and the derivative on the other side, well we need that exponential function its own derivative and the chain rule for that but then we get this. Now we have a close look and you see two bits that are the same which means we can also write X down here. Now divide and we’ve got one of those super famous facts from calculus, that the derivative of log X is 1/X. Wonderful! And now I’m just going to go for it. So integrate and we get this guy here. We substitute e for X and that gives this guy here. Now on the right side we’ve got log e and obviously that’s equal to 1 since our logarithm is base e. Almost there ! The standard geometric interpretation of this definite integral here is that the area under the graph of 1/X between 1 and e is exactly equal to 1. Now many people are under the impression that, unlike pi the number e does not have a nice geometric interpretation. Well here we’ve got one, right, in terms of the hyperbola which is both the graph of 1/X and a basic geometric shape closely related to pi’s circle. This is the way to remember it: open up this yellow curtain here under the hyperbola until it’s area is exactly one and you’ve found e. All right, now to wrap things up I’ll quickly sketch how you can also use our infinite sum identity to derive e to the i pi is equal -1 and that’s the standard way this is done in calculus. There are also ways to write the sine and cosine functions as infinite sums like this. After substituting i X for X in the top identity and multiplying the sine identity in the middle by i we’ve got exactly the same stuff in the purple and the blue boxes. Now since we’re dealing with identities, the same has to be true on the left. So the contents of the purple box is equal to the sum of what’s in the blue box, and this is Euler’s formula, super famous, right? Now I go over all this slowly in this other Simpson based video up there. So if you, you know, want it a bit slower, check it out. After plugging in pi for X the right side becomes -1 which then brings us full circle back to Euler’s identity. If you’ve made it this far you’ve really earned yourself the Mathologer’s extra huge seal of approval. So, as usual, I’m very interested in finding out how well all these explanations work for you. So please let me know in the comments. And that’s all for today.

100 thoughts on “The number e explained in depth for (smart) dummies”

  1. Here's something that's been bugging me: Why is (1+1/x)^x = e when x approaches infinity?shouldn't it be 1+1/inf)^inf at the limit, which would basically be 1? Because 1/x when x->inf is 0, and (1+0)^inf should be then why is it e and not 1?

  2. Absolutely awesome explanation. Everything that we just took for granted regarding e has been explained so very well here. Thanks for this lovely video.

  3. I wrote a C++ program to estimate e^x with that summation series.
    This program estimates Euler's number (e) to 10 decimal places if no more than 10 summations are added up,
    and if more estimates are used the precision is increased to 18.

    I think I can confidently estimate e^(x) to 18 decimal places if I add up 1000 terms from n = 0 to n = 1000.

    Here is the program:

    #include <iostream>
    #include <iomanip>
    #include <cmath>
    using namespace std;

    long double factorials(int n);

    int main()
    int x, maximum, precision;
    long double total = 0;

    cout << "This program will estimate the e^xnusing the infinite series: x^n / (n!) from n = 0 to large number nn";
    cout << "For any user defined value of x" << endl;

    cout << "Please enter a value of x. Use 1 if you want to estimate Euler's number: ";
    cin >> x;
    cout << "Please enter the span of your search from n = 0 to some number: ";
    cin >> maximum;

    if(maximum >= 0 && maximum <= 10)
    precision = 10;
    else if(maximum > 10)
    precision = 18;

    for(int n = 0; n < maximum; n++)
    total += pow(x,n) / (factorials(n));

    cout << "The estimated total of e^(x) over " << maximum << " iterations is " << fixed << setprecision(precision) << total << endl;

    return 0;

    long double factorials(int n)
    long double factorial = 1;
    if(n == 0)
    return factorial = 1;
    else if(n == 1)
    return factorial = 1;
    else if(n > 1)
    for(int i = 1; i <= n; i++)
    factorial *= i;
    return factorial;

    // Running the program I estimate Euler's number to 20 decimal places by adding up 1000 terms.

    This program will estimate the e^x
    using the infinite series: x^n / (n!) from n = 0 to large number n
    For any user defined value of x
    Please enter a value of x. Use 1 if you want to estimate Euler's number: 1
    Please enter the span of your search from n = 0 to some number: 1000
    The estimated total of e^(x) over 1000 iterations is 2.718281828459045235

    Process returned 0 (0x0) execution time : 5.487 s
    Press any key to continue.

  4. I like that episode when Homer has the pencil removed from his sinus and becomes a genius. I think the pencil was putting pressure on his brain. I think he started as a genius when he was a kid but a pencil got shoved up his nose then he became a dummy.

  5. excellent graphics. Just suggest people go to full screen at the start of the video, so they can get the benefit of the graphics.

  6. Hi Mr. O'Loger, what I think you really need is a video on i^i.  These explanations of e^ipi are all similar, and it's easy to get so you can completely internalize how to picture e^ipi without, I think, really really understanding it at all.  i^i exposes this weakness.  What is REALLY going on here?

  7. So is it possible to get that on a channelin german? I think i can hear some german accent (not bad) and I am german too. So is it possible to get those videos in german?

  8. Around this time in the video it is stated that the error is equal to something else. Was that a correct statement?

  9. Look at self roots, that is the nth root of n:1 root 1=1, 2 root 2 = 1.414…, 3 root 3 = 1.442…, 4 root 4 = 1.414…,  5 root 5 = 1.380…, 100 root 100 = 1.047… 3 has the biggest self root of any integer, but the number with the biggest of all is – yes – e, at 1.444667861…I found this by myself, which doesn't necessarily make it a new discovery of course. I'm only a smart dummy after all. But please tell me if you know of it anywhere else in the literature.(Another fact that's news to me is that the self root of 2 and that of 4 are equal. They seem to be the only two numbers sharing a self root. Right?)

  10. Well if that isn't Mr. "making maths entertaining again!" 🙂

    The way you explain the math of e feels like eating the spinich of the famous Popey the Sea-Man!

    You create the desire to also want to go deep down into other (hopefully still hidden) secrets of the quenn of sciences! It feels very motivating because you empower/reactivate the inborn intuition of people, where Einstein told us: "Thge only thing that really matters is intuition!" You make math feel come alive again and making it usefull for millions of people! Thank you for that valuable contribution to society!

    I'm actually a colleague of yours but not so much into software. If you tell me which software to use, then, once I become successfull with it, I'll mention you as the man who gave me the oportunity and connect with you, so we can both help even more people to inspire their lives with maths! 🙂

  11. I prefer the other proof of cis(x)=cos(x)+i sin(x)=e^ix

    Let f(x)=cis(x), note that f(0)=1
    f'(x) = -sin(x)+i cos(x) = if(x)
    i = f'(x)/f(x)
    Integrating both sides and using standard u-substition,
    We get,
    ix + c = log(f(x)) => f(x) = e^(ix+c) and since we know that f(0) = 1, we get c = 0. Therefore, e^ix = cis(x)

  12. Fantastic video as usual! Keep up the good work. For some reason, I find your videos more explanatory/fun than those of Numberphile (though I like them too!)

  13. Excellent! A great presentation. I look forward to viewing your other lectures. Also, The Simpsons can be a very educational show… if one pays attention; thanks for referencing the show. I’ve recorded it for years [decades] just so I can ‘rewind’ or freeze frame an interesting/educational reference.

  14. A big Mathematician, can't to remain in a logical slavery of Conventional, I hope at the last you understand what it mean. #Math #Mathematics #Science #News #ScienceNews #EarthProudDay Fermat's Last theorem is FUNDAMENTAL in Math and Science and after 380 years we can say in Culture.Fermat-Murgu Impossible Equations Sent Fermat's Last Theorem in FUNDAMENTAL By two Methods – One need a Postulate and I hope soon a Youngest Mathematician will be coming soon with , is about Fermat-Murgu n Media for 3 Integers associated to Fermat Equations – but second method is Definitive, and Second Grade Impossible Equations are absolute Conditionals for General Cases of n as Power and not only for 3 , 4, 5 and so on. See it at :" "

  15. So… why do math teachers say that pi can't be written as a fraction?
    With this identity, it clearly can be written as pi = ln(-1)/i …

  16. "e" is the elemental rate of change, the fundamental constituent of all rates, it's the dynamic existence of specific uncertainty, (where is the source/orign Singularity positioning, in infinite/eternal possibility?), containing the rational-numerical relative rates of exponential-logarithmic change, ..that combined, is the Universal differentially-integrated wave-package @ 1-0D singularity. Ie existence is here-now in actuality, and everywhen-there in the sum of all history probability, so if e-Pi-i is the singularity-source of continuous containment connection, then e^Pi-i is the "empty" return to Origin by reflection, because +1 -1= f(0) remainder 1/1 = 1, all history probability difference, in infinite closed-continuous possibility.
    It's the symmetrical phase-state of QM-Time.

    In terms of math-philosophy, "e" is "numberness".., virtual-mathematical connection, and the base self-defining quantity, .dt-quanta, of the eternally-infinite spectrum-quality, QM-TIME.., by instantaneous reflection-connection.
    (The "dumb" amateur explanation for smart people)

    Admire the "Mathematical Magic".., the continuing science of existence beyond rational comprehension…, n!->holistic completeness, as n->infinite.

    Covering the "sizeless" Singularity vanishing point location with infinite probability positioning points in Superposition = Superspin.
    This is the self-defining, (because e is the derivative of itself, existing in temporal superposition), observable mechanism, of QM-Time modulation in Quantum Field Operation, = Eternal Iteration = Temporal Substantiation of infinite probability in possibility = +/- f(1) connection.., and is equivalent to the concepts proposed by Professor John Wheeler's "One Electron" theory in these assembled mathematical components.

  17. Not only this is super neatly explained, it also includes The Simpsons and Katy Perry. Best video ever!

  18. I would go to 1 million and 1 digit, make a quick test to know if I should continue to 1 million and 2 digit, quick test, and so on, as long as my quick "carry test" proves that there is a doubt about the 1 million's digit value (and maybe some of its predecessors)'s value.




    Using a 120-centimetre length diameter line

    1. Multiply the 120-centimeters diameter line by 3
    2. The Circles length is 360-centimeters
    3. Every Circle has 360-degrees The Circles length is 360-centimeters, and each degree is 1-centimeter in length.
    4. Multiply the 120 centimetres diameter line by 4, and the square of diameter line is 480-centimeters, and the circle's length is three-quarters of this length.


    Reference Encyclopedia Britannica
    Quote: Archimedes (c. 250 BC) took a major step forward by devising a method to obtain pi to any desired accuracy, given enough patience. By inscribing and circumscribing regular polygons about a circle to obtain upper and lower bounds, he obtained 223/71 < π < 22/7, or an average value of about 3.1418.

    To cite this page:
    • MLA Style: "pi." Encyclopædia Britannica. Encyclopaedia Britannica 2008 Ultimate Reference Suite. Chicago: Encyclopædia Britannica, 2008.
    • APA Style: pi. (2008). Enc

    Reference Oxford English Dictionary
    1. The boundary which encloses a circle. 2. The distance around something.
    Note the two words encloses and around

    1. A line marking the limits of an area.
    Note the word line

    The numerical value of the circumference of a circle to its diameter (approximately 3.14i59)
    Note the word approximately

    In Sum

    It is clear from these definitions the original Archimedean formula (which was later given the name Pi and decimalized) of 22/7 as an improper fraction or 3 whole units of length plus one-seventh remaining, relates purely to the length of a line that circumnavigates around the area of a circle, to form a. *approximate boundary that separates the area of a circle from its surrounding area.

    Therefore it follows that regardless of all beliefs and claims to the contrary, all of the calculation’s concerning circles and curvature’s that have been carried out using the formula Pi are approximates, and as such are mathematically inaccurate.

    In nature rather than mankind’s linear geometry, shapes do not possess outlines or boundaries, a prime example of this is the circular shape of the Moon; the Moon does not have any form of outline but rather its light yellow colour is contrasted and silhouetted against the dark night sky.

    This is as with the shapes of all things, the atoms of the surfaces of their physical forms merge with the atoms of their surrounding environment e.g. the Earth’s atmosphere. And as such the closest we can get to accurately calculating the area of a circle is by use of the length of a circles edge, rather than inaccurate linear approximates.


    Using a 120-centimetres Diameter line

    Multiply the 120-centimeters diameter line by 3

    The Circle is 360-centimeters long

    Multiply the 360 centimeters diameter line by 360 = 129, 600 square-centimeters

    Divide 129, 600 square-centimeters by 12 = 10, 800 square-centimeters to the circles area


    Proposition 1.
    The area of any circle is equal to a right-angled triangle in which one of the sides about the triangle is equal to the radius, and the other to the circumference of the circle.

    The diameter line we are going to use to find the area of a circle according to Archimedes proposition 1, has a 120-centimetre length

    The triangles base-line is 60-centimeters as per the radius of the circle

    The height of the right angle is 360-centimeters as per the length of the circle

    60-centimeters base-line x 360-centimeters circle length yields a rectangle with an area of 21, 600 square-centimetres, which when divided by 2, yields an area of 10, 800 square-centimetres to the area of each triangle and the 120 centimetres diameter line circle.


    Using a 120-centimetre diameter line multiply it by itself to yield a square with an area of 14, 400 square-centimetres.

    Divide the square into four quadrants and each quadrant will have an area of 3, 600 square centimetres

    Divide one of the 3, 600 square centimetres quadrants by 4, and each of the four sub-quadrants will have an area of 900 square centimetres

    Multiply the area of three of the 900 square centimetre sub-quadrants (2.700 sq cm) by 4 = 10, 800 square centimetres to the area of the circle and three-quarters of the area to 14, 400 sq cm square

    The 60-centimetre radius of the circle’s diameter, gives 3, 600 square-centimetres of the area to one quadrant of the overall square.

    SUMERIAN METHOD 1000 BC; 10,800 square-centimetres to the circle

    ARCHIMEDES TRIANGLE 212 BC; 10,800 square-centimetres to the circle

    THREE TIMES THE RADIUS SQUARED 2017 AD; 10,800 square-centimetres to the circle

    FOUR QUADRANTS 2017 AD; 10, 800 square-centimetres to the circle

    FOUR IDENTICAL RESULTS, cannot be coincidental, and all four identical results are equally and self-evidently correct and incontestable.



    Calculating the surface area and volume of a 6-centimetre diameter sphere, obtained from a 6-centimetre cube.

    1. Measure the height of the cube to obtain a length of diameter line = 6 cm

    2. Multiply 6 cm diameter line by 6 cm to obtain the square area of one face of the cube; and also add them together to obtain the length of the perimeter to the square face = Length 24 cm, Square area 36 sq cm.

    3. Multiply the square area, by the length of diameter line to obtain the cubic capacity = 216 cubic cm.

    4. Divide the cubic capacity by 4, to obtain one-quarter of the cubic capacity of the cube = 54 cubic cm.

    5. Multiply the one quarter cubic capacity by 3. to obtain the cubic capacity of the Cylinder = 162 cubic cm.

    6. Multiply the area of one face of the cube by 6, to obtain the cubes surface area = 216 square cm.

    7. Divide the cubes surface area by 4, to obtain one-quarter of the cubes surface area = 54 square cm.

    8. Multiply the one quarter surface area of the cube by 3, to obtain the three quarter surface area of the Cylinder = 162 square cm.


    9. Divide the Cylinders cubic capacity by 4, to obtain one-quarter of the cubic capacity of the Cylinder = 40 & a half cubic cm.

    10. Multiply the one quarter cubic capacity by 3, to obtain the three quarter cubic capacity of the Sphere = 121 & a half cubic cm, to the volume of the Sphere.

    11. Divide the Cylinders surface are by 4, to obtain one-quarter of the surface area of the Cylinder = 40 & a half square cm.

    12. Multiply the one quarter surface area by 3 to obtain the three quarter surface area of the Sphere = 121 & a half square cm, to the surface area of the Sphere.

  20. The number of required terms for computing e to 10^6 decimal places is 205022.

    If we let f(n) to be the number of required terms for computing e to n decimal places, for 10^7 decimal places (and beyond) we have:

    f(10^7) = 1723507
    f(10^8) = 14842906
    f(10^9) = 130202808
    f(10^10) = 1158787577
    f(10^11) = 10433891463
    f(10^12) = 94851898540
    f(10^13) = 869200494599
    f(10^14) = 8019346203785
    f(10^15) = 74419210652835

    NOTE: f(n) can be computed very efficiently by using the binary search algorithm.

    See also:

  21. Ummmm, you need continuous convergence to be able to derive a function by deriving the sum of its component functions term by term, but I get why you didn't discuss that 😛

  22. e is irrational log is rational phi is circle I is imaginary plus or minus one and a zero. Part 90% solutions to include most number type. Transcendental including would be better.

  23. Hey can you use your maths to create a huge magnetic suit that would allow me to walk on the streets and suck up cars onto the suit.

  24. At about 14:39, when you are proving that the derivative of e is its self. You are using circular logic, because that series of "e" is derived from the fact that the derivative of "e" is "e".

  25. After 30 years since I studied calculus this has been the most BRILLIANT explanation of natural logarithms EVER ! FANTASTIC I WILL NEVER FORGET THIS 👍🏻GORGEOUS VIDEO

  26. i have a question about 5:30, how do they all go to 1? the last few would go to zero wouldnt they? because you'd have (x!)/(x^x) or close to it which goes to zero as x goes to infinity eg (1000!)/(1000^1000)= 4*10^-433

  27. Hi, Thank you for your videos. I have a question regarding the end result of your equation at 17:59. Is it possible to input numerical values into that equation and get the left side of the equation equals the right side of the equation? How do you do that? Thank you.

  28. Say I take an error of 1/3!, therefore (1+1/1!+1/2!+1/3!). If I now estimate 8/3 as e: 8/3 – (1+1/1!+1/2!+1/3!) = 0 < 1/7!; Am I missing something here?

  29. integral(sin^4t/cos^8t-4(sin^5t/cos^7t)+4(sin^6t/cos^6t))dt = u^4(1+u^2)du – 4u^5du + 4tan^6(t)dt, set u = sin(t)/cos(t) and finally I get 22/7 – pi

  30. Thanks you very much. I have been banging my head for math theories like this. 😳🙏🙏💓💕💖💖💗. Please recommend some books to know about facts like this . I will definitely follow them. Mainly to know about caluclus , trig derivation and complex, probability , binomial stuff .

  31. Ok so now i know how. But I still don't know really why. This is very annoying.

    This is the first of very many videos which made me understand how tho! Superb! Thanks!

    From what he says e^x =1+x+x^2/2!….
    So for x=π
    Am i wrong?

  33. Starting at 14:04. During any basic calculus courses, the fact that e^x is its own derivative is simply shoved down our throats. But after seeing the true explanation for it, I can now consider my mind thoroughly blown, and I can rest in peace.

  34. il y a les nombre cyclique aussi, ça fait 543210123456789X98765 et une histoire de lapin qui se reproduisent.

  35. Going through school, I noticed, that mathematical concepts stick in your head better, when you connect them to their actual, real life, application(s).

    All this fraction, (etc…), juggling is useless, if one does not understand/know where, how, and for what purpose IN REAL LIFE this, or that, concept is used.

    Dear Mathologer, perhaps, you could include an extra minute, or two, in all of your videos, with good examples of where, and how, the concepts you're talking about in that particular video apply.

    Without that, all your wonderful work is for nothing. Enters into one ear, leaves through the other, and the understanding of the need of this, or that, concept for real world use never sets in.

  36. at 5:26, why do those fractions turn to 1? If the number on the top gets smaller than the bottom one, shouldnt it go to 0 instead of 1?

  37. When I saw your shirt, I immediately wanted to prove the reslut. It turned out to be a lot easier than I thought 🙂

  38. I think there could be a misunderstanding between log and ln : shall we not use ln in place of log in this video?
    Of course ln(e) =1, and these are two differents functions.

  39. Primarily, via Mathologer, I've learned a lot about the concept of infinite sets. I like the proofs they can be employed to solve. However, doing deeper research, I'm finding that some professional Mathematicians don't believe in infinite sets. Can you clarify this battleground?

  40. Exponential series is summation integration- differentiation infinite series.

  41. very nice It's an approximation. that explain maybe why logarithm is approximation of harmonic number. Thank you very much I often have trouble understanding the exponential and logarithm now I understand better, thank you.

  42. Is there a way of interpretation of e^(i*pi)=-1 where one can explain it's meaning by simplest terms like debt and loss and whatever, avoiding all the trigonometry, infinite series and general calculus identities which kind are bound to get to the same point in case you start with Cantor's theorem and Continuum hypothesis as a beginning for all? Hence, in reality, all these equations are various ways of representation of the same thing, but this "thing" is an absolutely fantastic and totally incomprehensible unity where two totally different irrational (infinite) numbers somehow interact with each other and one totally imaginary symbol so to say – in a very non-trivial way – to produce the most basic number we are absolutely sure about. So, again, is there a way of interpretation of e^(i*pi)=-1 by very simple everyday terms (like debt and loss and gain and time etc.)?

  43. (e × cbrt e × 5th root of e × 7th root of e and so on till infinity)/(sqrt e × 4th root of e × 6th root of eand so on till infinity ) = 2

  44. I've watched quite a few of the Mathologer series and they are consistently well done, … and without a doubt one of the best at explaining complex mathematical concepts. Wish that these had been around when I was in school.

  45. The natural logarithm of a number is its logarithm to the base of the mathematical constant e, . The natural logarithm of x is generally written as ln x , loge x, or sometimes, if the base e is implicit, simply log x. Just saying…

  46. If the million plus one-th digit is going to push the millionth digit up, we would want another zero to guarantee that the millionth digit remains the same.

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