Power series: x^3/3 + x^9/9 + x^15/15 .......

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SUMMARY

The discussion focuses on finding the function represented by the power series \( \frac{x^3}{3!} + \frac{x^9}{9!} + \frac{x^{15}}{15!} \). Participants identify a connection to the power series expansion of \( e^x \) and suggest differentiating the series multiple times to derive a differential equation. The key insight is that the sixth, twelfth, and eighteenth derivatives of the function \( f(x) \) equal \( f(x) \) itself, leading to the differential equation \( f^{(6)}(x) = f(x) \). The discussion hints at utilizing complex roots of unity to solve this equation.

PREREQUISITES
  • Understanding of power series and their expansions
  • Familiarity with factorial notation and its application in series
  • Knowledge of differential equations, specifically higher-order derivatives
  • Basic concepts of complex analysis, particularly roots of unity
NEXT STEPS
  • Study the power series expansion of \( e^x \) and its applications
  • Learn techniques for solving higher-order differential equations
  • Explore the concept of complex roots of unity and their implications in differential equations
  • Investigate the properties of Taylor series and their convergence
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Mathematicians, students studying advanced calculus, and anyone interested in the analysis of power series and differential equations.

Shubhan672
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Moved from technical forum section, so missing the homework template

Let x be a real number. Find the function whose power series is represented as follows: x^3/3! + x^9/9! + x^15/15! ...

I see that there is a connection to the power series expansion of e^x but am having difficulty finding the function.
 
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Shubhan672 said:
Moved from technical forum section, so missing the homework template
Let x be a real number. Find the function whose power series is represented as follows: x^3/3! + x^9/9! + x^15/15! ...

I see that there is a connection to the power series expansion of e^x but am having difficulty finding the function.
Notice that ##x^3 = (x^3)^1## and ##x^9 = (x^3)^3##. Does that give you any ideas?
 
Try differentiating the series multiple times and see if you can determine a differential equation it satisfies.
 
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vela said:
Try differentiating the series multiple times and see if you can determine a differential equation it satisfies.
f(x) =
1581222250175.png


f'(x) =
1581222301684.png


f''(x) =
1581234957297.png


I can see that the 6 th, 12th, 18th ... derivative of f(x) would be f(x) itself but am having difficulty understanding its relevance and am having difficulty proceeding from here,
 

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  • 1581222032075.png
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Shubhan672 said:
f(x) = View attachment 256852

f'(x) = View attachment 256853

f''(x) = View attachment 256862

I can see that the 6 th, 12th, 18th ... derivative of f(x) would be f(x) itself but am having difficulty understanding its relevance and am having difficulty proceeding from here,

##f^{(6)}(x) = f(x)## is a differential equation. Can you solve it?

Hint: complex roots of unity.
 
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