Ай бұрын
4 ай бұрын
5 ай бұрын
5 ай бұрын
7 ай бұрын
8 ай бұрын
9 ай бұрын
10 ай бұрын
Жыл бұрын
Пікірлер
@Serghey_83 5 күн бұрын
У тебя G похожа на греческую сигму σ ))
@Serghey_83 5 күн бұрын
Вау!! Экуейшен!)))
@uwogd 6 күн бұрын
Or! You can prove F is a gradient of a potential function and use FTC for Line integrals to prove its 0 :)
@Mion_7 7 күн бұрын
Nice video
@KartikeyPatel-f8e 8 күн бұрын
Her equation ki ek limit hoti he bad me vo galat sabhit hote he kyoiki ye sirf ek mathematic hi he or kuch nahi hota he 😅vastvikata se sambandh rakhana chahiye kalpana se nahi 😅😅😅
@KartikeyPatel-f8e 8 күн бұрын
Her equation ki ek limit hoti he bad me vo galat sabhit hote he kyoiki ye sirf ek mathematic hi he or kuch nahi hota he 😅equation se duniya chalati nahi he 😅vastvikata se sambandh rakhana chahiye kalpana se nahi 😅partikal ka astitva hota he 😅partikal ek hakat he 😅dimension ka nahi 😅partikal ko dimension he dimension se partikal nahi 😅partikal se duniya chalati he 😅dimension se nahi 😅partikal ka bhavatik rup se astitva hota he 😅baki Sam hamare man ka bram 😅
@KartikeyPatel-f8e 8 күн бұрын
We'll information good show and 😅😅
@nicovulpus 14 күн бұрын
If we get the same answer, why use implicit differentiation?
@kamalmohamed7750 21 күн бұрын
Ok
@apriparentesi Ай бұрын
Sorry, but this video is totally wrong.The problem is that to find out what the derivative of a logaritm is, you have to know the value of the limit itself. So, to prove your limit, you cannot use any result that depends on the knowing of the limit itself, this means you canno use De l'Hopital. This is called circular reference and cannot be used as a proof. Please delete this video as fast as you can.
@Rk-bq5ic 2 ай бұрын
This is so great!!
@jardah38 3 ай бұрын
The boy is too impetuous. I don't think he explains anything. He's just showing off.
@hungnguyen-mi5hn 4 ай бұрын
??? you use the definition of e to proof that definition?;)))) bernoulli defined e is the limit of that function when he had the problem of continuous compounding of interest
@hungnguyen-mi5hn 4 ай бұрын
acob Bernoulli discovered this constant in 1683, while studying a question about compound interest:[5] An account starts with $1.00 and pays 100 percent interest per year. If the interest is credited once, at the end of the year, the value of the account at year-end will be $2.00. What happens if the interest is computed and credited more frequently during the year? If the interest is credited twice in the year, the interest rate for each 6 months will be 50%, so the initial $1 is multiplied by 1.5 twice, yielding $1.00 × 1.52 = $2.25 at the end of the year. Compounding quarterly yields $1.00 × 1.254 = $2.44140625, and compounding monthly yields $1.00 × (1 + 1/12)12 = $2.613035.... If there are n compounding intervals, the interest for each interval will be 100%/n and the value at the end of the year will be $1.00 × (1 + 1/n)n.[20][21] Bernoulli noticed that this sequence approaches a limit (the force of interest) with larger n and, thus, smaller compounding intervals.[5] Compounding weekly (n = 52) yields $2.692596..., while compounding daily (n = 365) yields $2.714567... (approximately two cents more). The limit as n grows large is the number that came to be known as e. That is, with continuous compounding, the account value will reach $2.718281828... More generally, an account that starts at $1 and offers an annual interest rate of R will, after t years, yield eRt dollars with continuous compounding. Here, R is the decimal equivalent of the rate of interest expressed as a percentage, so for 5% interest, R = 5/100 = 0.05.[20][21] cre:wikipedia
@Knight-nu3eu 4 ай бұрын
understood this WAY better than in class
@za_k3457 4 ай бұрын
Thanks goat
@nicolascamargo8339 4 ай бұрын
Genial la explicación
@gegebenein.gaussprozess7539 4 ай бұрын
Always the shame shit. It's not Yuler's number, its Oyler's. Have you ever been in a math class?
@TheCalcSeries 4 ай бұрын
Chill buddy... Many countries outside the U.S. pronounce "Euler" as "Yuler", especially hispanic ones. I first learned about him in a math class in Spanish, so that's how I pronounce it. I believe pronunciation is a minor detail in a video focused on math, so I'd recommend you focus on that instead.
@nicolascamargo8339 4 ай бұрын
Genial el tema y la explicación
@nicolascamargo8339 4 ай бұрын
Wow genial
@nicolascamargo8339 4 ай бұрын
Muy clara la explicación excelente
@khruschkukurudza 5 ай 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 5 ай бұрын
Wow, very nice. ❤❤❤❤❤
@ShanBojack 5 ай 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 6 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 5 ай бұрын
❤
@archangecamilien1879 5 ай 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 6 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 6 ай бұрын
❤
@MeyouNus-lj5de 6 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 6 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 6 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
@kubakopcil9992 4 ай бұрын
Your definition is only one of the few possible definitions of e. Another useful one is that the derivative of natural logarithm is the function 1/x . This video basically proves the equivalency of the two (rather, it proves that the first one follows from the second one, but the other direction isn't too different). Just a note: Logarithm is well definite no matter what, so we can use the fact that it's inverse function to exponential.
@uggupuggu 6 ай бұрын
nga how do you fine ln(x) without knowing e
@Bertin-q3y 6 ай бұрын
e^2
@solcarzemog5232 6 ай бұрын
Circular reasoning. You cannot use ln (log base e) trying to define e itself.
@holyshit922 6 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 6 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 7 ай бұрын
Very good. Thanks 👍
@secretsecret1713 8 ай бұрын
This is ridiculous. It is the definition of e.
@samueldeandrade8535 8 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 8 ай бұрын
This is NOT just a way to define e. This is the ORIGINAL way.
@finhz 8 ай бұрын
cool!
@charlespaxson2679 9 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 9 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 10 ай бұрын
Thanks bro i got it
@person1501aaa 10 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. 10 ай бұрын
good video man !! and thank for replying my mail on that day 👍👍👍👍
@TheCalcSeries 10 ай бұрын
Any time!
@pranjalsingh9784 11 ай бұрын
Gud job!!!🎉
@gibbogle Жыл бұрын
"Euler" is not pronounced you-ler, it is pronounced oiler. HTH
@victorfong8396 Жыл бұрын
it helpful!
@djttv Жыл 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 6 ай бұрын
Its a very obvious fact, but i dont know how to prove it 😂
@adw1z 6 ай бұрын
If f is continuous and lim f(x) exists, then the statement follows immediately
@binaryblade2 Жыл бұрын
Whole lotta math to avoid 2m = n
@trnfncb11 Жыл 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 Жыл 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 8 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...