Maclaurin series and general calculus question

In summary, a Maclaurin series is a special case of a Taylor series expansion used to approximate a function as a polynomial around x = 0. It can be calculated using a specific formula and is useful for simplifying complicated functions. It is also related to Taylor series, but with a point of expansion at x = 0. However, not all functions can be represented by a Maclaurin series, only those that are infinitely differentiable at x = 0.
  • #1
NihalRi
134
12

Homework Statement


This question has four parts which may follow up from each other so I incuded all the parts. The real problem I'm having is with d

Consider the function f ang g given by f (x)=( e^x+[e^-x])/2 & g (x) =( [e]^x]-[e^-x])/2
a) show f'(x) = g (x) and g'(x) = f (x)

b) find the first three non zero terms in the Maclaurin expansion of f (x)

c) hence find the value of lim (as x approaches 0) (1-f (x))/[x^2]

d) find the value of the improper integral ∫(0 to ∞) g (x)/[f (x)^2]dx

Homework Equations


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
images?q=tbn:ANd9GcQL15bOK9lgkXUTxBp16drYY1W4wb8wKLQfboepWV5Mwv0cDJsQvmRq5T-f.png


The Attempt at a Solution


a) was pretty straight forward using the quotient rule
b) using the formula I got
1 + [x^2]/2! + [x^4]/4!
c) I replaced c for f (x) and simplifying gave me -1/2
d) I replaced g (x) with its differential from a) becoming
∫ (0 to ∞) f'(x)/[f(x)^2]
= lim (b approaches ∞) ∫ (0 to b) f'(x)/[f(x)^2]
I thought maybe I'd try substitution now but that's not working
 
Last edited:
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  • #2
NihalRi said:

Homework Statement


This question has four parts which may follow up from each other so I incuded all the parts. The real problem I'm having is with d

Consider the function f ang g given by f (x)=( e^x+[e^-x])/2 & g (x) =( [e]^x]-[e^-x])/2
a) show f'(x) = g (x) and g'(x) = f (x)

b) find the first three non zero terms in the Maclaurin expansion of f (x)

c) hence find the value of lim (as x approaches 0) (1-f (x))/[x^2]

d) find the value of the improper integral ∫(0 to ∞) g (x)/[f (x)^2]dx

Homework Equations


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
images?q=tbn:ANd9GcQL15bOK9lgkXUTxBp16drYY1W4wb8wKLQfboepWV5Mwv0cDJsQvmRq5T-f.png


The Attempt at a Solution


a) was pretty straight forward using the quotient rule
b) using the formula I got
1 + [x^2]/2! + [x^4]/4!
c) I replaced c for f (x) and simplifying gave me -1/2
d) I replaced g (x) with its differential from a) becoming
∫ (0 to ∞) f'(x)/[g'(x)^2]
= lim (b approaches ∞) ∫ (0 to b) f'(x)/[f(x)^2]
I thought maybe I'd try substitution now but that's not working

What substitution did you try? There is one particular substitution that makes this problem very easy, indeed.
 
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  • #3
Ray Vickson said:
What substitution did you try? There is one particular substitution that makes this problem very easy, indeed.
Letting u = f (x) I think is ideal but my upper and lower limits would have to change to f (b) and -1 respectively. I'm getting,
∫(-1 to f (b)) u^-2 du
Which is [-1×u^-1](from -1 to f (b))
Since f (x) = u
We've got [-1×f(x)^-1)](from -1 to f (b))
now when I started replacing things started to get messy (I'd type it out but it's just really messy and probably won't end up making much sense) which makes me think that perhaps there is another way, that relates to part c) somehow.

I just tried making b approach infinity and I think the upper part of the definite integral vanishes and I'm left with just the lower part which equals 0.65. I'm not sure if this is right and I'm still suspicious that there could be another method:)
 
Last edited:
  • #4
NihalRi said:
Letting u = f (x) I think is ideal but my upper and lower limits would have to change to f (b) and -1 respectively. I'm getting,
∫(-1 to f (b)) u^-2 du
Which is [-1×u^-1](from -1 to f (b))
Since f (x) = u
We've got [-1×f(x)^-1)](from -1 to f (b))
now when I started replacing things started to get messy (I'd type it out but it's just really messy and probably won't end up making much sense) which makes me think that perhaps there is another way, that relates to part c) somehow.

I just tried making b approach infinity and I think the upper part of the definite integral vanishes and I'm left with just the lower part which equals 0.65. I'm not sure if this is right and I'm still suspicious that there could be another method:)
When you switch from u back to ƒ(x), the limits of integration go back to 0 and b .
 
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1. What is a Maclaurin series?

A Maclaurin series is a special case of a Taylor series expansion, where the function is approximated as a polynomial around the point x = 0. It is named after the Scottish mathematician Colin Maclaurin.

2. How is a Maclaurin series calculated?

A Maclaurin series can be calculated using the formula: f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ... + (f^(n)(0)x^n)/n!, where f^(n)(0) is the nth derivative of f at x = 0.

3. What is the purpose of a Maclaurin series?

Maclaurin series are useful for approximating complicated functions with polynomials, which are easier to work with. They are also used in calculus to find derivatives and integrals of functions.

4. How do Maclaurin series relate to Taylor series?

A Maclaurin series is a special case of a Taylor series, where the point of expansion is x = 0. This means that the coefficients in a Maclaurin series are calculated using derivatives evaluated at x = 0, while in a general Taylor series, the coefficients are calculated at a specific point x = a.

5. Can a Maclaurin series represent any function?

No, a Maclaurin series can only approximate functions that are infinitely differentiable at x = 0. This means that some functions, like piecewise functions with discontinuities at x = 0, cannot be represented by a Maclaurin series.

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