what's also interesting about this series is that it also works for some complex numbers. if you, for example, plug in x=i, you get i-1/2-i/3+1/4+i/5-...=log(1+i)=log(2)+i*pi/4, which is a nice way of proofing Leibnitz' series for pi/4.
This video show the actual algorithm used by computers to compute log functions. How a microprocessor which can only do the simplest primitive operations such as "+" and binary shift can be used to compute sin, log, etc? Power series is the answer. For those who wander why every engineer is forced to learn these concepts. It is because a few of them will want to compute things that no computer ever did before. Checking the convergence allow to find if the particular equation you selected will give the right answer for your set of data.
Certainly a computer implementation will use *a* power series ... but probably not this one. The reason is that limiting the number of terms will give an error (truncation error) which will increase as the argument increases. An approximation is required whose error is within uniform bounds over the entire range. The usual functions used are Chebyshev polynomials.
So if `x` can be only in the interval (-1,1], how can we get the power series for the remaining part of the graph of the `ln(1+x)` function? :q If you graph it, you can see that it does have valid values for `x > 1`, of course. So there should also be a way to express them with some power series, right? "Our friend" doesn't seem to give us the full picture then :q
There is no power series for log(1 + x) that converges for all x. The problem is that log(1 + x) is not an entire function; it hits a singularity at x = -1. Any series expansion for the logarithm, no matter where it is set, will eventually hit the singularity; when it does so determines the radius of convergence of the series. For example, the series for log(10 + x) has a radius of 10 because of the singularity at x = -10. The values for the logarithm beyond the radius of convergence are determined by different methods, including the definition of the logarithm as an integral or the inverse of the exponential function. Logarithm identities are also helpful, particularly the fact that log ab = log a + log b. Wikipedia explains that this property along with log tables were commonly used when logarithms were first introduced.
Well, every series with powers of _x_ are power series, and there are power series which do not correspond to any nice function in a closed form. Also, a Taylor series of a function is not neccessarily defined everywhere and one function can have different Taylor series depending on _x_ (or none at some points). But, if a function is analytical, its power series and its Taylor series are actually the same.
A power series is any series consisting of non-negative integer powers of (x - c) with each of them multiplied by a real constant (in other contexts, we can allow x and the coefficients to be complex, for example), where c is real number (or complex) called the "center" of the power series. It may converge or not. If it converges, we can find the so called "radius of convergence", say, R, which is a positive number. If R is the radius of convergence, we guarantee that a real power series centered on point c will converge at least on the open interval (c - R, c + R). A Taylor series is a particular power series where the coefficients are given by a formula that uses information about a function f and its derivatives to expand it in an infinite series. It's used to get an exact power series representation of the function f on some neighbourhood of the power series's center. Sometimes, the Taylor series representation of a function converges to the function on all the real line. In other times, just on some distance around the center. It may also not converge on any open interval at all (convergence on the center point is always guaranteed because the series equals zero)! It may happen as well that the Taylor series indeed converges in some interval, but not to the function we wanted to expand! The theory of Taylor series is very subtle and elegant, and there are many things to explore. Anyway, a quick answer is: Taylor series are a particular case of a general power series! :)
So then, why the input of the Zeta Funcition of (-1) is equal to (-1/12)?, because i have seen a video of numberphile in wich they prove this by arguing that 1-1+1-1....=(1/2), then that 1-2+3-4+5...=(1/4) but non of them shouldn´t converge, one being the case of the best friend when x=-1 and the other shouldn,t because of the AST.
Because the zeta function is a meromorphic continuation of the sum that you are familiar with. Hence, you may assign values where the zeta 'sum' does not converge, but preserves nice properties (such as analyticity). The theory for analytic/meromorphic continuation is quite different from that of power series.
@@iraklidiasamidze4400 Since the 0 in the exponent is an integer, my intuition would say that in this context of the sigma sum notation, 0^0 would give 1. But I agree that in general, 0^0 is undefined.
4050johnny ah true good catch. But since the interval of convergence is (-1,1]. Thus ln|1+x| is the same as ln(1+x) on that interval. Just for power series, the abs value was not needed but I should have mentioned that in the vid
I think you're wrong in that -1 doesn't work. ln(1+x) = that sum That sum turned out to be -infinity for x = -1 ln(1-1) = -inf ln(0) = -inf 0 = e^(-inf) = 1/e^(inf) = 1/inf = 0 0=0 so -1 works just fine
The thing is that we want to know where the series converges (that's why we calculated radius and integral of convergence), so at x=-1 we see it diverges.
That's what happens when you don't pay attention to concepts and just memorise applications, ln(0) is not "equal to" -infinity♾️. It is undefined, and ♾️ is not a number it is a concept. Which comes up when you talk about limits
what's also interesting about this series is that it also works for some complex numbers. if you, for example, plug in x=i, you get i-1/2-i/3+1/4+i/5-...=log(1+i)=log(2)+i*pi/4, which is a nice way of proofing Leibnitz' series for pi/4.
This video show the actual algorithm used by computers to compute log functions.
How a microprocessor which can only do the simplest primitive operations such as "+" and binary shift can be used to compute sin, log, etc? Power series is the answer.
For those who wander why every engineer is forced to learn these concepts. It is because a few of them will want to compute things that no computer ever did before. Checking the convergence allow to find if the particular equation you selected will give the right answer for your set of data.
Certainly a computer implementation will use *a* power series ... but probably not this one.
The reason is that limiting the number of terms will give an error (truncation error) which will increase as the argument increases. An approximation is required whose error is within uniform bounds over the entire range. The usual functions used are Chebyshev polynomials.
(7:36 ln(1+x=2) full integration) ,(9:33-12:36 harmonic series) (12:37- 14:48 how to change the index in the summation)
that was a deep sigh...
yea...
This is also the Maclaurin series for ln(1+x)!! 😄
Thanks is quite impressive teaching, I like it.
Really Thank you so much 🥺🙏💛
that's all, PANDA! thx for the vid
who knew this would help in a 2020 final calc exam
So if `x` can be only in the interval (-1,1], how can we get the power series for the remaining part of the graph of the `ln(1+x)` function? :q If you graph it, you can see that it does have valid values for `x > 1`, of course. So there should also be a way to express them with some power series, right? "Our friend" doesn't seem to give us the full picture then :q
There is no power series for log(1 + x) that converges for all x. The problem is that log(1 + x) is not an entire function; it hits a singularity at x = -1. Any series expansion for the logarithm, no matter where it is set, will eventually hit the singularity; when it does so determines the radius of convergence of the series. For example, the series for log(10 + x) has a radius of 10 because of the singularity at x = -10.
The values for the logarithm beyond the radius of convergence are determined by different methods, including the definition of the logarithm as an integral or the inverse of the exponential function. Logarithm identities are also helpful, particularly the fact that log ab = log a + log b. Wikipedia explains that this property along with log tables were commonly used when logarithms were first introduced.
power series are not the same as taylor series right? if it's true, what is the difference?
have a nice day
Well, every series with powers of _x_ are power series, and there are power series which do not correspond to any nice function in a closed form. Also, a Taylor series of a function is not neccessarily defined everywhere and one function can have different Taylor series depending on _x_ (or none at some points). But, if a function is analytical, its power series and its Taylor series are actually the same.
thanxs for the answer
A power series is any series consisting of non-negative integer powers of (x - c) with each of them multiplied by a real constant (in other contexts, we can allow x and the coefficients to be complex, for example), where c is real number (or complex) called the "center" of the power series. It may converge or not. If it converges, we can find the so called "radius of convergence", say, R, which is a positive number. If R is the radius of convergence, we guarantee that a real power series centered on point c will converge at least on the open interval (c - R, c + R).
A Taylor series is a particular power series where the coefficients are given by a formula that uses information about a function f and its derivatives to expand it in an infinite series. It's used to get an exact power series representation of the function f on some neighbourhood of the power series's center. Sometimes, the Taylor series representation of a function converges to the function on all the real line. In other times, just on some distance around the center. It may also not converge on any open interval at all (convergence on the center point is always guaranteed because the series equals zero)! It may happen as well that the Taylor series indeed converges in some interval, but not to the function we wanted to expand! The theory of Taylor series is very subtle and elegant, and there are many things to explore.
Anyway, a quick answer is: Taylor series are a particular case of a general power series! :)
I have terrible doubt ! How do ln2 series exist even the geometric series is defined for |x|
@@jigglygamer6887 ook ! Please tell me bit clearly, if u have time !
@@jigglygamer6887 thanks i got it
Same after 3 years
Sir you are awesome
So then, why the input of the Zeta Funcition of (-1) is equal to (-1/12)?, because i have seen a video of numberphile in wich they prove this by arguing that 1-1+1-1....=(1/2), then that 1-2+3-4+5...=(1/4) but non of them shouldn´t converge, one being the case of the best friend when x=-1 and the other shouldn,t because of the AST.
Because the zeta function is a meromorphic continuation of the sum that you are familiar with. Hence, you may assign values where the zeta 'sum' does not converge, but preserves nice properties (such as analyticity). The theory for analytic/meromorphic continuation is quite different from that of power series.
Oh, thank you
So is the interchange of limits okay in this scenario?
In this case because of the uniform convergence of the geometric series
@4:26 you have C on both sides of = sign. they cancel.
However, that by itself would not prove C = 0
The problem with that is that you also have to subtract from ln (1+X) thus getting ln (1+x) - C
Best way.... Tnx
Wanna mention that our best friend's formula with sigma doesn't work when x=0 because we get 0^0 + 0^1 + ... , instead of 1 + 0^1 + 0^2 + ...
0^0 is not equal to 1
@@mouhebmanai Well, yes, that's what I just said
But the limit as x→0 of x^0 will give 1
@@hetsmiecht1029 as x approaches 0, 0^x = 0, 0^0 is either 0 or 1, I guess depends on context or something
@@iraklidiasamidze4400 Since the 0 in the exponent is an integer, my intuition would say that in this context of the sigma sum notation, 0^0 would give 1. But I agree that in general, 0^0 is undefined.
where tf did (-1)^n come from though
Ok, so how often does ln(X+k)/ln(X) converge?
why are you dividing ln(x + k) by ln x ?
Being a mathematician, I can divide but anything. I can even get away with dividing b zero, once I meet l’Hopital.
Isn't the integral of 1/(1+x) = ln|1+x| ? once we have R = 1 we can see that |1 + x| = 1 + x but maybe we needed it?
4050johnny ah true good catch. But since the interval of convergence is (-1,1]. Thus ln|1+x| is the same as ln(1+x) on that interval. Just for power series, the abs value was not needed but I should have mentioned that in the vid
Please prove that the alternated harmonic series converges
You can apply alternating series test to see this result.
AST
You surmise -1
i love you
What about ln(5-x)
Just replace x by 4-x
i want to be ninjaaaaa...welcome to china town.
Good
If you would have set limits then C would have automatically canceled out?
That's why you never need to worry about + C in a definite integral :)
Great!
How to do the best friend :((
It's from taylor series. And you get taylor series from induction.
T?HANK YOU
This series can be calculated by taking nth derivative instead of integrating geometric series
I think you're wrong in that -1 doesn't work.
ln(1+x) = that sum
That sum turned out to be -infinity for x = -1
ln(1-1) = -inf
ln(0) = -inf
0 = e^(-inf) = 1/e^(inf) = 1/inf = 0
0=0
so -1 works just fine
The thing is that we want to know where the series converges (that's why we calculated radius and integral of convergence), so at x=-1 we see it diverges.
That's what happens when you don't pay attention to concepts and just memorise applications, ln(0) is not "equal to" -infinity♾️. It is undefined, and ♾️ is not a number it is a concept. Which comes up when you talk about limits
:)