How Does the -1 Arise in This Series to Function Conversion?

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SUMMARY

The discussion centers on the infinite sum represented by the equation $$\sum_{n=1}^\infty \frac{MK^{n-1}|t-t_0|^n}{n!}$$ and its conversion to the function $$\frac{M}{K}(e^{K(t-t_0)}-1)$$. The key point of confusion is the appearance of the -1 in the function, which arises because the sum starts at n=1, omitting the constant term (1) from the series expansion of the exponential function. This missing term accounts for the -1 when the constant is moved from the right side to the left side of the equation.

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transmini
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I ran across an infinite sum when looking over a proof, and the sum gets replaced by a function, however I'm not quite sure how.

$$\sum_{n=1}^\infty \frac{MK^{n-1}|t-t_0|^n}{n!} = \frac{M}{K}(e^{K(t-t_0)}-1)$$

I get most of the function, I just can't see where the ##-1## comes from. Could someone help show that?
 
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transmini said:
I ran across an infinite sum when looking over a proof, and the sum gets replaced by a function, however I'm not quite sure how.

$$\sum_{n=1}^\infty \frac{MK^{n-1}|t-t_0|^n}{n!} = \frac{M}{K}(e^{K(t-t_0)}-1)$$

I get most of the function, I just can't see where the ##-1## comes from. Could someone help show that?

The first term in the expansion of ##e^x## is ##1##, usually put in the sum with index ##0##. In your case the sum starts with ##n=1## so the constant term is missing on the left side. If you move the constant term from the right side to the left you will see it.
 
LCKurtz said:
The first term in the expansion of ##e^x## is ##1##, usually put in the sum with index ##0##. In your case the sum starts with ##n=1## so the constant term is missing on the left side. If you move the constant term from the right side to the left you will see it.
Oh, not sure how I missed that. That makes sense, thanks
 

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