How to write this function in series form?

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Discussion Overview

The discussion revolves around expressing a function in series form, specifically focusing on the function f(n) involving terms with variables z and s. Participants explore how to generalize the representation of the series to avoid writing out numerous terms explicitly, particularly when n is large.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant asks how to express the function f(n) in a series form that stops adding terms when n equals x.
  • Another participant corrects the notation in the original post, clarifying that the terms should be z^(n-1), 100^(n-1), etc.
  • A participant expresses uncertainty about the terminology, questioning if "series form" is the correct term and seeking a more general way to write the function to simplify it for large n.
  • One participant proposes a general j^th term for the series as s*z^(n-j)/100^(n-j) and discusses the implications of a factor of 2 in the second term, questioning if it should apply to other terms as well.
  • The same participant reiterates the general term and emphasizes the need to consider the factor of 2, suggesting that it may require a separate addition if it is the only term with a different coefficient.
  • A later reply confirms the presence of the factor of 2 and inquires how to determine the number of terms needed for the function to equal 100,000 given specific values for z and s.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the best way to express the function in series form, and multiple viewpoints regarding the generalization and notation remain present.

Contextual Notes

There are unresolved questions about the correct application of coefficients in the series terms and the implications of the factor of 2 in the second term. Additionally, the discussion does not clarify the specific conditions under which the series should be truncated.

nickadams
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How can I write this is series form?

f(n) = [itex]\frac{z^n * s}{100^n}[/itex] + 2*[itex]\frac{z^(n-1) * s}{100^(n-1)}[/itex] + [itex]\frac{z^(n-2) * s}{100^(n-2)}[/itex] + [itex]\frac{z^(n-3) * s}{100^(n-3)}[/itex] + [itex]\frac{z^(n-4) * s}{100^(n-4)}[/itex]I stopped at this amount of terms arbitrarily, but really I want the series to stop adding terms when n is equal to x in "z^(n-x)" and "100^(n-x)". How can I say that in a general form?
 
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EDIT: it is supposed to be z^(n-1), 100^(n-1), z^(n-2) and 100^(n-2) etc..

not 100(n-1), 100(n-2) etc.. as it shows up in the OP
 
I don't even know if it is called series form, but I was just wondering if there was a more general way to write it? Like if n is equal to 100, I don't want to have to write out 100 terms of the equation so it would be simpler if there were a generalized form...
 
Well, the general j^th term looks like

[tex]s\frac{z^{n-j}}{100^{n-j}},[/tex]

so you'll just need to sum over the range of j you want. However, in your original post you had a 2 in your second term - is that is really supposed to be there? If yes, should there be similar factors in front of the other terms? If so you will need to modify the above general term accordingly. If the term with the two is the only one like that, then summing over the above general term will leave you short a [itex]sz^{n-1}/100^{n-1}[/itex], so you'd have to add that separately.
 
Mute said:
Well, the general j^th term looks like

[tex]s\frac{z^{n-j}}{100^{n-j}},[/tex]

so you'll just need to sum over the range of j you want. However, in your original post you had a 2 in your second term - is that is really supposed to be there? If yes, should there be similar factors in front of the other terms? If so you will need to modify the above general term accordingly. If the term with the two is the only one like that, then summing over the above general term will leave you short a [itex]sz^{n-1}/100^{n-1}[/itex], so you'd have to add that separately.

Thank you so much! And yes, the 2 was supposed to be there. So how can I find out how many terms it will take for my function in the OP to equal 100,000 if z=5 and s=500?
 

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