exp(x) = sum(x^n/n!,n=0..infinity) If we want to eliminate odd terms we can calculate 1/2(f(x)+f(-x)) 1/2(exp(x)+exp(-x)) So we have cosinus hiperbolicus cosh(1)
I knew the sum right away, special sums were on our last math analysis test, consider also expansions for (1±x)^a and other derived sums (through sum differentiation, integration and multiplying by x) for extra questions
Right out of the gate I observe the sum to be the even part of e^x evaluated at 1. That is, cosh(1) = 1/2(e^1 + e^-1) = (e^2 + 1)/2e. For those who don't know, cosh and sinh are by definition the even and odd parts of e^x respectively.
I'm going to be pedantic and say the use of "By definition" implies you also define e^x using the series expansion. There are several ways to define these functions, not just power series.
@@bsmith6276 that's the wrong thing to be pedantic about 😅. The power series is very arguably the best definition, and the one that's used for more exotic arguments
cos(x) has only even powers of x, so plugging in i will only yield real valued results. i^2n = (-1)^n. Here’s how you get the result: Plug in ix and -ix into the Taylor series of e^x, and compare it to the Taylor series of sinx and cosx. You’ll find that: e^(ix) = cosx + i sinx e^(-ix) = cosx - i sinx This can be used to derive the following: cos(x)=(e^(ix)+e^(-ix))/2 Simply plug in i as the input, and you’ll find that: cos i = (e+e^-1)/2.
Simple. That is the hyperbolic cosine chx := ½(e^x+e^-x) , it has the taylor expansion : Ch X = Σ 1/2n! .x^2n Obviously it is an even function and we have e^x = Ch X +shx. And for our special case u just put X=1 and u get the sum equals Ch 1.
@@robertveith6383 it is indeed. But do you have an application that adds a keyboard on smartphones for fast writing in mathematics? You know how much time it took me to write that.
This sum is also the Taylor expansion for cosh(1).
exp(x) = sum(x^n/n!,n=0..infinity)
If we want to eliminate odd terms we can calculate
1/2(f(x)+f(-x))
1/2(exp(x)+exp(-x))
So we have cosinus hiperbolicus
cosh(1)
I knew the sum right away, special sums were on our last math analysis test, consider also expansions for (1±x)^a and other derived sums (through sum differentiation, integration and multiplying by x) for extra questions
Right out of the gate I observe the sum to be the even part of e^x evaluated at 1. That is, cosh(1) = 1/2(e^1 + e^-1)
= (e^2 + 1)/2e.
For those who don't know, cosh and sinh are by definition the even and odd parts of e^x respectively.
I'm going to be pedantic and say the use of "By definition" implies you also define e^x using the series expansion. There are several ways to define these functions, not just power series.
@@bsmith6276 that's the wrong thing to be pedantic about 😅. The power series is very arguably the best definition, and the one that's used for more exotic arguments
Your last line is incorrect. Because of the Order of Operations, you need grouping symbols in the denominator: (e^2 + 1)/(2e).
Very nice.
Thank you
My intuition had me thinking 1.5 to 1.6. Just didn't know how to actually get there mathematically.
ch1
👍
(e + 1/e) /2
better known as cosh (1)
cos(i)?😮😮😮
yeah or cosh(1)
which equals cos(i)=cosh(1)=(e+1/e)/2. If you know what deratives are you pretty much derive what cos(i) is
Thank you
@@RR-bs9mrwhy??
cos(i*x) = cosh(x). If we let x=1, then cos(i) = cosh(1).
cos i = (e^2 +1)/(2e)
Why????
Exactly! Why? 😁
cos(x) has only even powers of x, so plugging in i will only yield real valued results. i^2n = (-1)^n.
Here’s how you get the result:
Plug in ix and -ix into the Taylor series of e^x, and compare it to the Taylor series of sinx and cosx. You’ll find that:
e^(ix) = cosx + i sinx
e^(-ix) = cosx - i sinx
This can be used to derive the following:
cos(x)=(e^(ix)+e^(-ix))/2
Simply plug in i as the input, and you’ll find that:
cos i = (e+e^-1)/2.
@@gregstunts347 I am convinced, thanks👍
cos(z) = (exp(iz) + exp(-iz))/2
plug in z = i
Simple. That is the hyperbolic cosine chx := ½(e^x+e^-x) , it has the taylor expansion :
Ch X = Σ 1/2n! .x^2n
Obviously it is an even function and we have e^x = Ch X +shx.
And for our special case u just put X=1 and u get the sum equals Ch 1.
nice!
Your second line is written wrong. Each place that has 2n must have grouping symbols around it. 1/(2n)! and x^(2n)
@@robertveith6383 it is indeed. But do you have an application that adds a keyboard on smartphones for fast writing in mathematics? You know how much time it took me to write that.
@@SyberMath
And by the way it is also the same error in the video.
Cosh-ерное видео ))