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@khruschkukurudza 6 күн бұрын
This proof is invalid because it contains circular reasoning (circulus in probando). Derivation of the formula for the derivative of natural logarithm, which is being used to find linear approximation is based on the very same limit which you are trying to prove. So implicatively you are using the limit in question as a part of its own proof.
@RSLT 6 күн бұрын
Wow, very nice. ❤❤❤❤❤
@ShanBojack 20 күн бұрын
While others are criticizing you, ill appreciate you for your efforts. Keep going you'll do well. Also only take constructive criticism. The ones who are just making fun of you are jerks. Dont be a d³x/dt³.
@archangecamilien1879 Ай бұрын
I think I remember it was supposed to be e^2 or something, lol...that generally, the limit of (1+x/n)^n as n goes to infinity is e^x...but that's just memory...or maybe false memory, lol...
@RSLT 6 күн бұрын
❤
@archangecamilien1879 6 күн бұрын
Perhaps one way to do this is to use L'Hôpital's rule, lol...take the limit of the natural log of that (there must be a theorem saying the natural log of the limit is the limit of the natural log, lol)...then get rid of that n up there, take the limit using L'Hopital's rule because it's an indeterminate form, etc...infinity x 0, etc...see what that yields...
@Marek_Wiesner Ай бұрын
Well just to make your result slightly simpler I would instead of for eg. 2sinˇ3(x)/3 write 1/3 2sinˇ3(x).
@kannankk2001 Ай бұрын
❤
@MeyouNus-lj5de Ай бұрын
Theorem 5: The Euler characteristic of a topological space is a topological invariant. Proof: Let X be a topological space, and let χ(X) be its Euler characteristic, defined as: χ(X) = Σ_i (-1)^i β_i where β_i is the i-th Betti number of X, which counts the number of i-dimensional "holes" in X. To prove that χ(X) is a topological invariant, we need to show that it remains unchanged under continuous deformations of X, such as stretching, twisting, or bending, but not tearing or gluing. Consider a continuous map f : X → Y between two topological spaces X and Y. The induced homomorphisms on the homology groups of X and Y satisfy the following property: f_* : H_i(X) → H_i(Y) is a group homomorphism for each i Moreover, the alternating sum of the ranks of these homomorphisms is equal to the Euler characteristic: Σ_i (-1)^i rank(f_*) = χ(X) - χ(Y) Now, if f is a homeomorphism, i.e., a continuous bijection with a continuous inverse, then the induced homomorphisms f_* are isomorphisms, and their ranks are equal to the Betti numbers of X and Y: rank(f_*) = β_i(X) = β_i(Y) for each i Therefore, we have: Σ_i (-1)^i rank(f_*) = Σ_i (-1)^i β_i(X) - Σ_i (-1)^i β_i(Y) = χ(X) - χ(Y) = 0 This implies that χ(X) = χ(Y) whenever X and Y are homeomorphic, i.e., χ is a topological invariant. This proof highlights the fundamental role of the Euler characteristic in capturing the essential topological properties of a space, and suggests that the concept of zero or nothingness may be intimately connected to the deep structure of space and time.
@KingGisInDaHouse Ай бұрын
And the award for the cutest math teacher goes to ...this guy. I thought Flammy went uncontested for a while lol. But yeah this is how I approached the problem as well but whether or not thats actually L'Hopitals rule with extra steps is debatable.
@adw1z Ай бұрын
e^x = lim n->inf (1+x/n)^n by definition, it’s when u start trying to “prove” it some other ways such as using log which is the inverse of e^x to prove e^x that u start running into circular argument problems so beware. Here you are good though, u didn’t run into any circular arguments which is impressive
@uggupuggu Ай бұрын
nga how do you fine ln(x) without knowing e
@user-nd7th3hy4l Ай бұрын
e^2
@solcarzemog5232 Ай бұрын
Circular reasoning. You cannot use ln (log base e) trying to define e itself.
@holyshit922 2 ай бұрын
I expected reduction formula derived by parts wyth Pythagorean trigonometric identity or Chebyshev polynomials if we want to stay in real domain We can also express this integral as sum
@holyshit922 2 ай бұрын
Yes , substitution is good and quite fast but reduction formula is not so bad either moreover reduction works for both odd and even cases of n
@surendrakverma555 2 ай бұрын
Very good. Thanks 👍
@secretsecret1713 3 ай бұрын
This is ridiculous. It is the definition of e.
@samueldeandrade8535 3 ай бұрын
Some people already commented, but it is good to make it clear: this proof is invalid. But it is also common. So this youtuber shouldn't feel bad at all.
@samueldeandrade8535 3 ай бұрын
This is NOT just a way to define e. This is the ORIGINAL way.
@finhz 3 ай бұрын
cool!
@charlespaxson2679 4 ай бұрын
An easy way to see that something is amiss is to use a logarithm to another base. Say use log base 10, then you’d prove that the limit equals 10, not e.
@lovishnahar1807 5 ай бұрын
i have doubt in my mind on this way of proving that this involves lo hopitals rule and taking log abe e itself , actually think there musty be algeraic way to arrive here which is based on more rigour, hope u helpme with this
@hassnbabar2955 5 ай бұрын
Thanks bro i got it
@person1501aaa 5 ай бұрын
i ahve been sitting here for an hour struggled on why the fact that 3^4x is (4)(3^4x)(ln3) and after searhcing and aksing people i finally saw this and the explaination is amazing
@lostInSocialMedia. 5 ай бұрын
good video man !! and thank for replying my mail on that day 👍👍👍👍
@TheCalcSeries 5 ай бұрын
Any time!
@pranjalsingh9784 6 ай бұрын
Gud job!!!🎉
@gibbogle 7 ай бұрын
"Euler" is not pronounced you-ler, it is pronounced oiler. HTH
@victorfong8396 7 ай бұрын
it helpful!
@djttv 7 ай бұрын
How do we prove that: ln(lim f(x)) = lim ln(f(x))? It seems intuitive, but I suppose it must be proved.
@btb2954 2 ай бұрын
Its a very obvious fact, but i dont know how to prove it 😂
@adw1z Ай бұрын
If f is continuous and lim f(x) exists, then the statement follows immediately
@binaryblade2 7 ай бұрын
Whole lotta math to avoid 2m = n
@trnfncb11 7 ай бұрын
Wait a minute. How do you know that the derivative of ln x is 1/x without knowing that the limit you started with is equal to e??
@TheCalcSeries 7 ай бұрын
Indeed, the proof that shows that the derivative of lnx is one over x already assumes the limit equals e. I didn't know this when I recorded this video, but as others say, this feels like circular reasoning!
@samueldeandrade8535 3 ай бұрын
@@TheCalcSeries this doesn't "feel like" circular reasoning. This IS circular reasoning. But that's ok. Pretty common in proofs involving e. Usually people make circular reasoning to prove the continuity or differentiability of e^x. Your video is NOT about those. You just make a little honest understandable confusion. The limit in the video is the ORIGINAL definition of e. It gives the expression e = 1/0! + 1/1! + 1/2! + ... To see that, notice that for fixed n, (1+1/n)^n = sum_k C(n,k) (1/n)^k = sum_k n!/(k!(n-k)!) (1/n)^k = sum_k n(n-1)...(n-(k-1))/k! 1/n^k = sum_k 1/k! n/n (n-1)/n ... (n-(k-1))/n = sum_k 1/k! 1 (1-1/n) ... (1-(k-1)/n) of course, k goes from 0 to n. For n big, each term 1-k/n approximates 1 and products of those terms also approxinate 1, so each term of the sum will be approximately 1/k!, which means lim (1+1/n)ⁿ = sum_k 1/k! Some particular examples: n=1 (1+1/1)¹ = 1+1 n=2 (1+1/2)² = 1+2(1/2)+(1/2)² = 1+1+1/4 = 2.25 n=3 (1+1/3)³ = 1+3(1/3)+3(1/3)²+(1/3)³ = 1+1+1/3+1/27 = 2.370370... n=4 (1+1/4)⁴ = 1+4(1/4)+6(1/4)² +4(1/4)³+(1/4)⁴ = 1+1+3/8+1/16+1/256 = 2.44140625 n=5 (1+1/5)⁵ = 1+5(1/5)+ 10(1/5)²+10(1/5)³ +5(1/5)⁴+(1/5)⁵ = 1+1+2/5+2/25+1/125+1/3,125 = 2.48832 n=6 (1+1/6)⁶ = 1+6(1/6)+ 15(1/6)²+20(1/6)³+15(1/6)⁴ +6(1/6)⁵+(1/6)⁶ = 1+1+5/12+5/54+5/1296+1/1296+1/6⁶ ≈ 2.5108239... It goes to e, but slowly. The obtained expression e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + ... goes faster! Indeed, n=2 sum_k 1/k! = 2.5 already n=3 sum_k 1/k! = 2.6666... n=4 sum_k 1/k! = 2.7083333... n=5 sum_k 1/k! = 2.716666... n=6 sum_k 1/k! = 2.718055555... n=7 sum_k 1/k! ≈ 2.7182539...
@miguelfelipe435 7 ай бұрын
The moment the limit becomes the derivative is worth it😃
@powercables1611 7 ай бұрын
Overall is ok but you should be more careful, for example at 5:50 this is not the definition of derivative of log because that requires taking the limit of a continuous variable. Here your delta x is a sequential limit. Luckily you are fine here because by archimedean principle you can find two continuous variable a,b such that a < 1/(2n) < b, and since ln(1+x)/x is stricly decreasing you can use the squeeze theorem to show that the limit at 5:50 is indeed equal to that if you had use a continuous variable for the limit
@deadpoolspatan1177 7 ай бұрын
You can do it by letting N = n/2, the limit then becomes : lim (1+1/N)^2N = lim ((1+1/N)^N)^2 with N going to infinity. And this is (lim (1+1/N)^N)^2 which equals e^2 as N goes to infinity
@btb2954 2 ай бұрын
thats what i did
@efegulcek9024 7 ай бұрын
luv u man you've been very helpful at my midterms
@ishrakmujibift4269 7 ай бұрын
Not to sound harsh, but you haven’t really shown anything in this video. You have shown that the limit of (1+1/n)^n as n approaches infinity is the number e. BUT THAT IS THE DEFINITION of e! How can you prove a definition? At the same time, you did not compute e, which you said you would do.
@Memories_broken_ 7 ай бұрын
He showed us how to compute the limit in the standard way rather than using the usual form which is exp(Limit x tends to (..) f(x)/g(x)) for the 1^infinity form
@Grim_Reaper_from_Hell 7 ай бұрын
There is only one way to define e -- e=limit (1+1/x)^x. Euler defined e as a Continuous Compounding of interest rates which is limit (1+i/x)^(tx)which iis equal e^(it). If you set i and t to 1 you get e. e is not an abstract concept but has real physical meaning. And he is trying to show that e = e
@Memories_broken_ 7 ай бұрын
Didn't Bernoulli do that?
@Grim_Reaper_from_Hell 7 ай бұрын
@@Memories_broken_ There are a lot of pretenders on the throne of been 1st. Euler is not one of them. But Euler was the one who named the constant in his own honour.😄😄😄😄 Even more interesting is that we are using the ln (base e) to prove that e equal to e
@Memories_broken_ 7 ай бұрын
@@Grim_Reaper_from_Hell oh no Bernoulli wasn't a pretender. He did notice the number appearing while working with interests and compounded sums, he just didn't dwell into it.
@Grim_Reaper_from_Hell 7 ай бұрын
@@Memories_broken_ Bernoulli is not the only one. If you look through the various books on the topic you will find a number of different names. Compounding was a popular topic back then and it is not surprising that a number of people derived the number before Euler and Bernoulli was one of them but was the Bernoulli 1st is not clear.
@samueldeandrade8535 3 ай бұрын
You are just 4wful. Have some decency.
@lostInSocialMedia. 7 ай бұрын
where i can talk to you ?
@TheCalcSeries 5 ай бұрын
How can I help you?
@Darkjnr5 7 ай бұрын
Thanks Sir
@shaimaasoltan5334 8 ай бұрын
kzbin.info/www/bejne/jXvMnJV-j9x-otksi=Jmk5z9V_luspvv3O
@TimoYlhainen 8 ай бұрын
Amazingly useless.
@derekjones4039 8 ай бұрын
I like the video but isn't this circular reasoning? You cannot say that the limit (the one where delta x approaches zero) is the derivative of ln(x) if the value of the limit is needed to prove the derivative of ln(x) at x=1.
@adventoabdielnababan1952 7 ай бұрын
Absolutely, i thought he was trying to calculate the value of “e”.😂
@jamesharmon4994 7 ай бұрын
@@adventoabdielnababan1952me too
@Hobbitangle 7 ай бұрын
Definitely it's a kind of circular proof. Or even a trivial proof like this: "Suppose that γ is equal to 'e'. So because the limit of the expression (1+1/n)^n is equal to γ and γ is equal to 'e', then the limit is also equal to 'e'. The author of the video is genius!
@phscience797 7 ай бұрын
While the author of this video certainly didn‘t elaborate that far, one can avoid circularity. For instance, you could define ln(x) as the antiderivative of 1/x.
@Hobbitangle 7 ай бұрын
@@phscience797 @phscience797 "you could define ln(x) as the antiderivative of 1/x" No, you couldn't. You may not use any derivatives of exp(x) or ln(x) till you prove the Second Wonderful limit.
@billprovince8759 8 ай бұрын
Nicely explained, but it doesn't match the description of the video! The video stated how to calculate e, which of course immediately conjures the Taylor series expansion for e^x evaluated at 1. Nevertheless, I appreciated your proof that this limit approaches e. .... Here's a related question you might want to consider: how well does this formula approximate e when n is some value like 10 or 100 or 1000. How would you characterize the convergence?
@edwardlitrenta5730 8 ай бұрын
How can a limit be a number?
@fylthl 8 ай бұрын
e isn't a number?
@billprovince8759 8 ай бұрын
If there was no limit (such as growing without bounds or simply never getting closer and closer to a single value), only then would you have to worry about a limit not being a number.
@abhilashpradhan7671 7 ай бұрын
@@fylthle is a constant. We need more reasoning for why we get e.
@nymph6282 8 ай бұрын
a function can be decreasing but be still positive so why does the derivative have to be positive?
@Karelpieter 7 ай бұрын
just in this case it is so
@Karelpieter 7 ай бұрын
if your case you wouldnt use mean value theorem
@fredsalter1915 8 ай бұрын
Future MIT professor here
@samueldeandrade8535 3 ай бұрын
Well, he is young, so he can be whatever he wants to if he works hard for it.
@eedestifan8047 8 ай бұрын
lhpotial make it way too easy
@meesveldstra2794 8 ай бұрын
Ty for the video mate, really helped me for my Calc finals :D
@marceloflores4870 8 ай бұрын
wonderful years KEVIN ARNOLD!!!!!!!
@christophernodurft1868 8 ай бұрын
This can be completely proven by substitution, trigonometry, and the Squeeze Theorem. If we let y=x-π/2, then x=y+π/2 and lim(x->π/2)=lim(y->0). Therefore, lim(x->π/2)(cos(x)/(x-π/2))= lim(y->0)(cos(y+π/2)/y)= lim(y->0)((cos(y)cos(π/2)-sin(y)sin(π/2))/y)= lim(y->0)((cos(y)(0)-sin(y)(1))/y)= lim(y->0)((0-sin(y))/y)= lim(y->0)(-sin(y)/y)= -lim(y->0)(sin(y)/y)= -1. The limit as we approach 0 for sin(y)/y is 1. This is well-known and can be proven by the Squeeze Theorem.
@guliyevshahriyar 8 ай бұрын
Teaching many concepts in 1 video. Thank you very much!