1. ## maclaurin series help

find the maclaurin series for f(x) = coshx

the answer is = 1 + x^2 / 2! + x^4 / 4! + x^6 / 6! + ...

my question: How is the 6th derivative positive? i believe its a -coshx?

also, whats the difference between coshx and just cosx?

this particular problem was unique to me. thanks

2. Originally Posted by rcmango
find the maclaurin series for f(x) = coshx

the answer is = 1 + x^2 / 2! + x^4 / 4! + x^6 / 6! + ...

my question: How is the 6th derivative positive? i believe its a -coshx?

also, whats the difference between coshx and just cosx?

this particular problem was unique to me. thanks
Third time lucky.

cosh(x) = [e^x+e^(-x)]/2

sinh(x) = [e^x-e^(-x)]/2

d/dx cosh(x) = sinh(x)

d/dx sinh(x) = cosh(x)

sinh(0) = 0

cosh(0) = 1.

From which the result follows from the definition of a Maclaurin series

(or even more easily by observing from the definition of cosh its power
series representation about 0 contains the even power terms of the
series for the exponential function, and the odd power terms all have
0 for their coefficient, and so don't appear in the expansion)

RonL

3. Originally Posted by rcmango
find the maclaurin series for f(x) = coshx

the answer is = 1 + x^2 / 2! + x^4 / 4! + x^6 / 6! + ...

my question: How is the 6th derivative positive? i believe its a -coshx?

also, whats the difference between coshx and just cosx?

this particular problem was unique to me. thanks

The hyperbolic cosine and sine are not sine and cosine functions at all, however they are called like that because they are very similar related through the derivatives. And furthermore the analogue of a sine and cosine on a hyperbolic geometry are that of a sinh and cosh x.
Many important identities can be remembered through them as thinking about them as sine and cosine. For example, the infinite series you gave seems like the sin x series however this one has positive ters and is therefore a hyperbolic sine.