MHB What is the correct power series for $g(t)$?

Click For Summary
The discussion centers on determining the correct power series for the function \( g(t) \). A proposed series, \( g(t) = \sum_{n=0}^{\infty}\frac{g(t)}{n!} \), is identified as incorrect, as it only represents the zero function. The correct formulation is clarified as the Taylor power series, expressed as \( g(t) = \sum_{n=0}^{\infty} \frac{(t-t_0)^n g^{(n)}(t_0)}{n!} \). This series uses derivatives of \( g \) evaluated at a specific point \( t_0 \). The conversation concludes with an acknowledgment of the initial misunderstanding regarding the series representation.
Dustinsfl
Messages
2,217
Reaction score
5
What would be the power series of $g(t)$?

$$
g(t) = \sum_{n=0}^{\infty}\frac{g(t)}{n!}
$$

This?
 
Physics news on Phys.org
?? What EXACTLY are you trying to do? I'm guessing the recursive definition you have suggested is not a good way to go.
 
dwsmith said:
What would be the power series of $g(t)$?

$$
g(t) = \sum_{n=0}^{\infty}\frac{g(t)}{n!}
$$

This?

There is a mistake in your post, as posted there is no such function other than the zero function.

CB
 
The power series of a function around a point \( t_0 \) is $$ g(t) = \sum_{n=0}^{\infty} \frac{(t-t_0)^n g^{(n)}(t_0)}{n!} .$$ Note that the \( g^{(n)}(t_0) \) denotes the derivative evaluated at \( t_0 \), where \( g^{(0)}(t_0) = g(t_0) \).

Does this help you?

Edit: Sorry, to be specific this is the Taylor power series.
 
There was a mistake in what I was reading, i.e. is read it wrong. I see what my issue was so the question I asked was wrong.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
View full post »

Similar threads

Undergrad On (real) entire functions and the identity theorem
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad What is the difference between these two power series theorems?
  • · Replies 5 ·
Replies
5
Views
2K
Graduate Solving a power series
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad On convolution theorem of Laplace transform: Schiff
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Question on an infinite summation series
  • · Replies 5 ·
Replies
5
Views
3K
MHB Find Power Series Representation for $g$: Interval of Convergence
  • · Replies 5 ·
Replies
5
Views
2K
MHB Power Series Convergence Assistance
  • · Replies 3 ·
Replies
3
Views
2K
MHB How Do You Evaluate the Indefinite Integral as a Power Series?
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Taylor Expansion Question about this Series
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Looking for references on this form of a Taylor series
  • · Replies 3 ·
Replies
3
Views
2K