An Interesting Sum With Reciprocals of Factorials

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[@Blaqjaqshellaq]

This sum is also the Taylor expansion for cosh(1).

[@seanfraser3125]

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.

[@bsmith6276]

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.

[@KeimoKissa]

​@@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

[@robertveith6383]

Your last line is incorrect. Because of the Order of Operations, you need grouping symbols in the denominator: (e^2 + 1)/(2e).

[@holyshit922]

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)

[@Michael-cg7yz]

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

[@guts1859]

My intuition had me thinking 1.5 to 1.6. Just didn't know how to actually get there mathematically.

[@jaimeduncan6167]

Very nice.

[@SyberMath]

Thank you

[@honestadministrator]

(e + 1/e) /2 better known as cosh (1)

[@giuseppemalaguti435]

ch1

[@AbouTaim-Lille]

👍

[@AbouTaim-Lille]

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.

[@SyberMath]

nice!

[@robertveith6383]

Your second line is written wrong. Each place that has 2n must have grouping symbols around it. 1/(2n)! and x^(2n)

[@AbouTaim-Lille]

@@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.

[@AbouTaim-Lille]

@@SyberMath And by the way it is also the same error in the video.

[@Nobodyman181]

cos(i)?😮😮😮

[@RR-bs9mr]

yeah or cosh(1)

[@RR-bs9mr]

which equals cos(i)=cosh(1)=(e+1/e)/2. If you know what deratives are you pretty much derive what cos(i) is

[@Nobodyman181]

Thank you

[@yusufdenli9363]

​@@RR-bs9mrwhy??

[@stephenshefsky5201]

cos(i*x) = cosh(x). If we let x=1, then cos(i) = cosh(1).

[@yusufdenli9363]

cos i = (e^2 +1)/(2e) Why????

[@SyberMath]

Exactly! Why? 😁

[@gregstunts347]

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.

[@yusufdenli9363]

@@gregstunts347 I am convinced, thanks👍

[@adw1z]

cos(z) = (exp(iz) + exp(-iz))/2 plug in z = i