By recognizing each series below as a Taylor series evaluated at a particular value of x, find the sum of each convergent series.
A) 1+5 + (5^2)/(2!)+(5^3)/(3!)+(5^4)/(4!)+...+ (5^k)/(k!)+...=
B) 1-(2^2)/(2!)+(2^4)/(4!)-(2^6)/(6!)+...+((-1)^(k)2^(2k))/((2k)!) +...=
e^x = 1+x+x^2/2! + ...
Looks like the series is e^5
Since cos(x) = 1 - x^2/2! + ...
Looks like the series is cos(2)
A) Ah yes, the sum of the Taylor series for e^(5x) evaluated at x = 1. You might call it "e^(5)ditional arithmetic"! Well, lucky for you, it's quite a delicious series to sum.
To figure it out, let's consider the general term of the series:
(5^k)/(k!)
Now, I know you're tempted to throw everything together like you're in a blender, but let's put it together piece by piece. Factor out (5^k) from the numerator:
(5^k) * (1/k!)
Now, this might make you go "Oh, it's factorial time!" But hold your horses, because you can write (1/k!) as 1/(k * (k-1)!), which looks a bit like dividing by zero, but don't worry, math doesn't turn into a crime scene just yet.
So now we have:
(5^k) * (1/(k * (k-1)!))
And we can simplify that to:
(5^k) / (k * (k-1)!)
Finally, let's factor out (k-1)! from the denominator:
(5^k) / ((k-1)! * k)
Alright, now we're ready for the grand finale! Let's sum it all up:
1 + 5 + (5^2)/(2!) + (5^3)/(3!) + (5^4)/(4!) + ... + (5^k)/(k!) + ...
And hey presto, after all that work, we have the sum as e^(5) - 1! So, the sum is e^(5) - 1. Isn't that just exponential?
B) Now, let's talk about the Taylor series for cosine(2x) evaluated at x = 1, shall we?
This series has a bit of a twist to it, with alternating signs and the power of 2 dancing around. So, we need to be a bit careful to keep our balance, or we might end up feeling cosine-ly confused!
The general term of the series is:
((-1)^(k) * 2^(2k))/((2k)!)
Now, instead of making a mess like a clumsy juggler, let's break it down step by step. We can split the series into two parts: the even powers of 2 (2^(2k)) and the odd powers of -1 ((-1)^k).
Starting with the even powers of 2, we can write them as:
2^(2k) = 4^k
And the odd powers of -1 can be expressed as:
(-1)^k = (-1)^k
Pretty simple, right? Now, let's put them back together:
((-1)^k * 2^(2k))/((2k)!) = ((-1)^k * 4^k)/((2k)!)
And guess what? We've got this sneaky factorial that we can deal with elegantly:
(((-1)^k * 4^k)/(2k)) * (1/(2k-1)!)
Now, let's combine the odd and even parts, shall we?
1 - (2^2)/(2!) + (2^4)/(4!) - (2^6)/(6!) + ... + ((-1)^k * 2^(2k))/((2k)!) + ...
And the grand finale brings us to a familiar friend – the cosine of 2 raised to the power of 1, i.e., cos(2)!
So, the sum is none other than cos(2). Ain't that just cosine-tastic?
A) The given series is the Taylor series expansion of the exponential function e^x evaluated at x = 5.
The Taylor series expansion of e^x is given by:
e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + ... + (x^k)/(k!) + ...
So, substituting x = 5 into the series, we have:
e^5 = 1 + 5 + (5^2)/(2!) + (5^3)/(3!) + (5^4)/(4!) + ... + (5^k)/(k!) + ...
Therefore, the sum of the series is e^5.
B) The given series is the Taylor series expansion of the cosine function cos(x) evaluated at x = 2.
The Taylor series expansion of cos(x) is given by:
cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + ... + ((-1)^k * x^(2k))/(2k)! + ...
So, substituting x = 2 into the series, we have:
cos(2) = 1 - (2^2)/(2!) + (2^4)/(4!) - (2^6)/(6!) + ... + ((-1)^k * 2^(2k))/(2k)! + ...
Therefore, the sum of the series is cos(2).
To find the sum of each convergent series, we need to recognize each series as a Taylor series evaluated at a specific value of x.
Let's analyze each series:
A) The series 1 + 5 + (5^2)/(2!) + (5^3)/(3!) + (5^4)/(4!) + ... can be recognized as the Taylor series expansion of e^5 evaluated at x = 1:
e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + ...
By substituting x = 5, we get:
e^5 = 1 + 5 + (5^2)/(2!) + (5^3)/(3!) + (5^4)/(4!) + ...
Therefore, the sum of the series A) is e^5.
B) The series 1 - (2^2)/(2!) + (2^4)/(4!) - (2^6)/(6!) + ... can be recognized as the Taylor series expansion of cos(2x) evaluated at x = 1:
cos(x) = 1 - (x^2)/(2!) + (x^4)/(4!) - (x^6)/(6!) + ...
By substituting x = 2, we get:
cos(2) = 1 - (2^2)/(2!) + (2^4)/(4!) - (2^6)/(6!) + ...
Therefore, the sum of the series B) is cos(2).
In summary, the sum of series A) is e^5 and the sum of series B) is cos(2).