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ssd
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Homework Statement
S = 1+ x/1! +x2/2! +x3/3! +...+xn/n!
To find S in simple terms.
Homework Equations
None
The Attempt at a Solution
I tried with Taylor's expansion, coshx and sinhx expansions. But cannot see consequence.
As far as I know, all you can get is ##e^x -\sum_{k=0}^n \frac{x^k}{k!} = r_n(x)## with some boundaries ##c \le r_n(x) \le C##ssd said:Homework Statement
S = 1+ x/1! +x2/2! +x3/3! +...+xn/n!
To find S in simple terms.
Homework Equations
None
The Attempt at a Solution
I tried with Taylor's expansion, coshx and sinhx expansions. But cannot see consequence.
According to Maple, the sum can be expressed in terms of an incomplete Gamma function ##\Gamma(n+1,x)## and some other factors, but I am not sure you would call that "simple".ssd said:Homework Statement
S = 1+ x/1! +x2/2! +x3/3! +...+xn/n!
To find S in simple terms.
Homework Equations
None
The Attempt at a Solution
I tried with Taylor's expansion, coshx and sinhx expansions. But cannot see consequence.
Just to check, S isn't given like this, is it?ssd said:S = 1+ x/1! +x2/2! +x3/3! +...+xn/n!
To find S in simple terms.
No, only n+1 terms.Mark44 said:Just to check, S isn't given like this, is it?
S = 1+ x/1! +x2/2! +x3/3! +...+xn/n! + ...
Those already look like simple terms to me.ssd said:Homework Statement
S = 1+ x/1! +x2/2! +x3/3! +...+xn/n!
To find S in simple terms.
s=exssd said:Homework Statement
S = 1+ x/1! +x2/2! +x3/3! +...+xn/n!
To find S in simple terms.
Homework Equations
None
The Attempt at a Solution
I tried with Taylor's expansion, coshx and sinhx expansions. But cannot see consequence.
No: absolutely not! The infinite sum is ##e^x## but--at least in the initial post--the OP is asking about the finite sum, just for the first ##n+1## terms of the exponential series.coolul007 said:s=ex
An exponential series is a mathematical series in which each term is multiplied by a constant number, called the common ratio, to get the next term. The general formula for an exponential series is a + ar + ar^2 + ar^3 + ..., where a is the first term and r is the common ratio.
The sum of (n+1) terms in an exponential series is given by the formula Sn = a(r^(n+1)-1)/(r-1), where Sn is the sum of the first n+1 terms, a is the first term, and r is the common ratio.
To find the sum of (n+1) terms in an exponential series, use the formula Sn = a(r^(n+1)-1)/(r-1), where Sn is the sum of the first n+1 terms, a is the first term, and r is the common ratio. Simply plug in the values for a, r, and n+1 and solve for Sn.
Yes, the sum of (n+1) terms in an exponential series can be infinite if the common ratio r is greater than 1. In this case, the series will continue to grow without bound as n increases, resulting in an infinite sum.
The sum of (n+1) terms in an exponential series is important in understanding the behavior and growth of the series. It can also be used in various applications, such as calculating compound interest or population growth. Additionally, it can help determine if a series is convergent (has a finite sum) or divergent (has an infinite sum).