Series Expansion: Show Sin^Cos x = x + O(x^3)

In summary, a series expansion is a mathematical tool used to approximate a function using a finite number of terms. It is useful for functions that are difficult to integrate or differentiate. The purpose of expanding Sin^Cos x is to obtain a more simplified expression for the function, making it easier to work with. The expansion is also helpful for understanding the behavior of the function and making predictions about its values. The notation "O(x^3)" in the series expansion indicates the order of the terms, with terms beyond the third term being negligible for accurate approximation. However, the series expansion can only provide an approximation of Sin^Cos x and will never give an exact value, even with more terms added.
  • #1
polygamma
229
0
Show that for small positive $x$, $$\left( \sin x \right)^{\cos x} = x -\left( 3 \log x + 1\right) \frac{x^{3}}{3!} + \Big( 15 \log^{2} x + 15 \log x + 11 \Big) \frac{x^{5}}{5!} + \mathcal{O}(x^{7})$$
 
Last edited:
Mathematics news on Phys.org
  • #2
$$ \large (\sin x)^{\cos x} = e^{\cos (x) \log (\sin x) } $$

$$ \large = e^{\cos (x) [\log (x- \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \ldots)]}$$

$$ \large = e^{\cos (x) [\log x + \log (1-\frac{x^{2}}{3!} + \frac{x^{4}}{5!} + \ldots)]}$$

$$ \large = e^{\cos (x) [ \log x -(\frac{x^{2}}{3!} - \frac{x^{4}}{5!} + \frac{x^{4}}{2(3!)^{2}} + \ldots)]}$$

$$ \large = e^{\cos (x) (\log x - \frac{x^{2}}{3!} - \frac{x^{4}}{180} + \ldots)} $$

$$ \large =e^{(1- \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \ldots )(\log x - \frac{x^{2}}{3!} - \frac{x^{4}}{180} + \ldots )}$$

$$ \large = e^{\log x - \frac{x^{2}}{3!} - \frac{x^{4}}{180} - \frac{x^{2}}{2!} \log x + \frac{x^{4}}{2!(3!)} + \frac{x^{4}}{4!} \log x + \ldots}$$

$$ = \large x e^{-\frac{x^{2}}{3!} - \frac{x^{2}}{2!} \log x} e^{\frac{7x^{4}}{90}+ \frac{x^{4}}{4!} \log x} \times \cdots$$

$$ =x \left( 1 - \frac{x^{2}}{3!} - \frac{x^{2}}{2!} \log x + \frac{1}{2!} \left(\frac{x^{2}}{3!} + \frac{x^{2}}{2!} \log x \right)^{2} + \ldots \right) \left( 1 + \frac{7 x^{4}}{90} + \frac{x^{4}}{4!} \log x + \ldots \right) \times \cdots$$

$$ = x \left(1 + \frac{7x^{4}}{90} + \frac{x^{4}}{4!} \log x - \frac{x^{2}}{3!} - \frac{x^{2}}{2!} \log x + \frac{x^{4}}{2!(3!)^{2}} + \frac{2 x^{4}}{(2!)^{2}(3!)} \log x+ \frac{x^{4}}{2!(2!)^{2}} \log^{2} x + \ldots \right) $$

$$ = x - x \left( \frac{x^{2}}{3!} + \frac{x^{2}}{2!} \log x \right) + x \left(\frac{11 x^{4}}{120} + \frac{x^{4}}{8} \log x + \frac{x^{4}}{8}\log^{2} x \right) + \ldots $$

$$ = x - \left( 3 \log x +1 \right) \frac{x^{3}}{3!} + \left(15 \log^{2} x + 15 \log x + 11 \right) \frac{x^{5}}{5!} + \ldots $$
 
Last edited:

Related to Series Expansion: Show Sin^Cos x = x + O(x^3)

What is a series expansion?

A series expansion is a mathematical tool used to approximate a function using a finite number of terms. It is particularly useful for functions that cannot be easily integrated or differentiated.

What is the purpose of expanding Sin^Cos x?

The purpose of expanding Sin^Cos x is to obtain a more simplified and manageable expression for the function. This can help with calculations and analysis of the function.

Why is the expansion of Sin^Cos x useful?

The expansion of Sin^Cos x is useful because it allows us to approximate the function with a polynomial, which is a much simpler form to work with. It also helps us to understand the behavior of the function and make predictions about its values.

What does the notation "O(x^3)" mean in the series expansion?

The notation "O(x^3)" represents the order of the terms in the expansion. It indicates that the terms in the expansion become increasingly smaller as x approaches 0, and the terms beyond the third term can be neglected without significantly affecting the overall accuracy of the approximation.

Can the series expansion be used to find the exact value of Sin^Cos x?

No, the series expansion is only an approximation of Sin^Cos x. As more terms are added to the expansion, the approximation becomes more accurate, but it will never be an exact representation of the function.

Similar threads

Replies
3
Views
821
  • General Math
Replies
5
Views
575
  • General Math
Replies
33
Views
2K
  • General Math
Replies
1
Views
682
Replies
2
Views
1K
Replies
4
Views
509
  • General Math
Replies
11
Views
1K
Replies
3
Views
1K
Replies
5
Views
869
Replies
1
Views
672
Back
Top