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The Summing template on the C Key how does it work?
05-11-2019, 11:54 AM
Post: #1
The Summing template on the C Key how does it work?
[attachment=7229]Product: HP Prime G2
I tried using The Sum that E for lists with i=0 and infinity sign above and the expression after the = sign.

I put the template on my Home page and added all the parts it did not work.

Does anyone know how to use it

I have tried it in CAS and without CAS still, this function does not operate for me.
Help needed.
[Image: 7116522]

https://h30434.www3.hp.com/t5/Calculator...-p/7116522

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05-11-2019, 01:12 PM
Post: #2
RE: The Summing template on the C Key how does it work?
The lower-case (script) i is the constant \( \sqrt{-1} \); use n for your index variable.

Graph 3D | QPI | SolveSys
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05-11-2019, 01:21 PM
Post: #3
RE: The Summing template on the C Key how does it work?
1. (Lower case) i is a reserved variable,
2. The angle mode should be set to radians,
3. Use == for the comparison.
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05-11-2019, 02:57 PM (This post was last modified: 05-11-2019 03:06 PM by tom234.)
Post: #4
RE: The Summing template on the C Key how does it work?
Thank you I have done the following, however, the answer is 1.9 It should be
1.5 for pi/2 and the answer should be 1.

I can't do this in a function app. I had to do it in just the regular window under CAS.
Is that where I should be entering it.
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05-12-2019, 08:47 AM
Post: #5
RE: The Summing template on the C Key how does it work?
It works fine in CAS.


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Josep Mollera. HP PRIME, HW: C, SW: 2.1.14730 (2023 04 13).
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05-12-2019, 10:50 AM
Post: #6
RE: The Summing template on the C Key how does it work?
(05-11-2019 02:57 PM)tom234 Wrote:  Thank you I have done the following, however, the answer is 1.9 It should be
1.5 for pi/2 and the answer should be 1.

I can't do this in a function app. I had to do it in just the regular window under CAS.
Is that where I should be entering it.

The problem with what is shown in your thumbnail is that the \((2n+1)!\) in the denominator is outside the summation, which is why it still appears in the final answer.

Nigel (UK)
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