Deriving e from the limit (1+1/x)^x as x approaches infinity

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Deriving e from the limit (1+1/x)^x as x approaches infinity

Рет қаралды 22,029

Tambuwal Maths Class

Күн бұрын

Пікірлер: 36

@Ursus-tl5lt 8 ай бұрын

You cant use ln in your proof since ln is based on e and its derivative is based on the results of this theorem. The actual proof is complicated and it just proves the limit exists and is between 2 and 3, than it is calculated with taylor approximation

@ntlake 7 ай бұрын

Not necessarily. You can define ln(x) without using e at all.

@blackapple62 7 ай бұрын

@@ntlake How exactly can you do this? Using the natural logarithm quite literally means you're using a logarithm of base e.

@ntlake 7 ай бұрын

@@blackapple62 well, in more advanced mathematics the natural logarithm log(x) is often defined as the integral from 1 to x of dt/t. Then you prove that it's also the inverse function of exp(x).

@lemon.linguist 6 күн бұрын

@@ntlakeexactly you can do it by figuring it out as an area function or antiderivative of 1/x

@isabellamichalek7705 Ай бұрын

hello, thank you for your video, but I am confused about how you canceled the -1/x^2.

@jamesharmon4994 4 ай бұрын

What I'd be interested in knowing is what happens if 1/x is replaced by 2/x, or 3/x. Since the given formula converges and one factor is being multiplied by a constant, the new formula should also converge.

@Osirion16 Жыл бұрын

Very cool thank you

@tambuwalmathsclass Жыл бұрын

Thank you

@РусскийПатриотЯша Ай бұрын

Idk, there was a n easier method imo, where by using log properties and mclaurin you not only simplify the process but also save up time.

@thedeathofbirth0763 6 ай бұрын

Awesome explanation!

@prakashlakhapate1598 3 ай бұрын

You can not use natural logarithm as E is to be proved. Dhanyavad

@El0melette Ай бұрын

you can if you define lnx as the integral from 1 to x of 1/t dt, and define e as the point where ln(x)=1.

@sweetworldismine Жыл бұрын

I just discovered your channel. You are doing a wonderful job. Do you have a Whatsapp or telegram group for Jss1 student's mathematics

@tambuwalmathsclass Жыл бұрын

I appreciate it 🙏 We only have a group on Facebook

@upalsengupta5878 4 ай бұрын

Excellent

@DamienBlt Жыл бұрын

I understood all, but at the beguining, why there is (let y) what meaning ?, and what ln y = lim.....

@lubanlatif3713 Жыл бұрын

Basically, he assigned "y" to the equation as a variable. Therefore, we assumed that "y" equaled the whole equation. In the case of (ln y = lim), we multiplied both sides with "ln" (excluding the limit). Later, he explained it in more detail.

@DamienBlt Жыл бұрын

@@lubanlatif3713 ok thanks !

@adw1z 8 ай бұрын

ln == log base e , and e is defined by this limit, so nothing was achieved other than showing consistency - there is no need to prove anything, as the definition of e itself is the limit

@ntlake 7 ай бұрын

Nope. e isn't defined by this limit, it's defined by the limit as n approaches infinity of (1+1/n)ⁿ

@adw1z 7 ай бұрын

@@ntlake that’s the exact same thing?? Using n,x,α,γ,.. whatever letter u want doesn’t matter as it is a dummy variable And if u mean n as in natural number or discrete limit, it doesn’t matter as the limiting function is continuous anyways, the discrete and continuous limits must agree

@ntlake 7 ай бұрын

@@adw1z yeah, but no. The definition of e is the limit of a sequence, then it being equal to the limit of the function is something you have to prove. Anyway, I can define e without using this limit if you want.

@adw1z 7 ай бұрын

@@ntlake the proof follows trivially directly from the definition of continuity of a function, but I’d be interested in seeing your other definition

@ntlake 6 ай бұрын

@@adw1z sorry, missed the notification. the fact that the proof is trivial doesn't change the fact that it isn't its definition. Anyway the other definition is e = 1/0! + 1/1! + 1/2! + 1/3! + ...

@prakashlakhapate1598 3 ай бұрын

Binomial expansion formula need to be used because e is to be proved.

@AbhilashKhuntia Жыл бұрын

I am getting stuck like why are we cancelling -1/x^2 won't those two be equal to 0 and we cannot cancel 0 terms like that Please correct me if I am wrong

@nychan2939 10 ай бұрын

You can cancel equal non-zero factors from the numerator and denominator. They may be small but mustn't be zero. You can't do the cancellation if both are zero. This is the main idea in limit evaluation.