Recursive Function Question PLEASE HELP

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Tekilou

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Could anyone please help on answering this question especially the second part of the question. Thanks very much I really appreciate it.20220219_235603.jpg
 
Tekilou said:
Could anyone please help on answering this question especially the second part of the question. Thanks very much I really appreciate it.View attachment 31216
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It seems you have solved the first of your problem. Please share your solution to (i).

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
Sn = 1 if n= 1
3S(n-1) if n >1

Thats my answer for part i) not sure if it's correct. But I'm confused how to do part ii) could u answer it for me. Thanks
 
Tekilou said:
Sn = 1 if n= 1
3S(n-1) if n >1

Thats my answer for part i) not sure if it's correct. But I'm confused how to do part ii) could u answer it for me. Thanks
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[math]\text{Day 1}= 1 = 2^0\\ \text{Day 2}= 2^0\cdot2=2^1\\ \text{Day 3}= 2^1\cdot2= 2^2\\ \text{Day 4}=2^2\cdot 2=2^3\\ \vdots\\ \text{Day n-1}=?\\ \text{Day n}=?\\[/math]Do you see a pattern?
 
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Tekilou said:
Sn = { 1 if n = 1
S(n-1) + 2^S(n-1) if n >1
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What you presented is not an equation as there's no equal sign. I don't know why you're eager to get to part ii), when you can't get part i) right. If you can't get part i) right then you're not gonna get part ii) right. They're related. Double-check your answer, make sure it is an equation and it makes sense.
Suppose I want to know how many were sick on day 4 i.e. n=4, and I use what you presented I get S(4-1)+2^S(4-1)=2^2+2^(2^2). Does that make sense?
 
Tekilou said:
Sn = 1 if n= 1
3S(n-1) if n >1

Thats my answer for part i) not sure if it's correct. But I'm confused how to do part ii) could u answer it for me. Thanks
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Your answer is correct, If [imath]D(1)=1[/imath] is the number infected on day one then
for [imath]n>1[/imath] we have [imath]D(n)=2\cdot D(n-1)+D(n-1)=3\cdot D(n-1)=3^{n-1}[/imath] is the number infected on day [imath]n[/imath].
SEE HERE Scroll down to see the table of values.
Now as to the second part I don't think it can be done without better information about rates of recovery.
[imath][/imath]
 
pka said:
Your answer is correct, If [imath]D(1)=1[/imath] is the number infected on day one then
for [imath]n>1[/imath] we have [imath]D(n)=2\cdot D(n-1)+D(n-1)=3\cdot D(n-1)=3^{n-1}[/imath] is the number infected on day [imath]n[/imath].
SEE HERE Scroll down to see the table of values.
Now as to the second part I don't think it can be done without better information about rates of recovery.
[imath][/imath]
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Suppose n=2, D(2)=3^(2-1)=3. That's not true.
Edit: My mistake, you're right.
 
Thank you for your help
pka said:
Your answer is correct, If [imath]D(1)=1[/imath] is the number infected on day one then
for [imath]n>1[/imath] we have [imath]D(n)=2\cdot D(n-1)+D(n-1)=3\cdot D(n-1)=3^{n-1}[/imath] is the number infected on day [imath]n[/imath].
SEE HERE Scroll down to see the table of values.
Now as to the second part I don't think it can be done without better information about rates of recovery.
[imath] [/QUOTE][/imath]
Click to expand...
 
BigBeachBanana said:
Suppose n=2, D(2)=3^(2-1)=3. That's not true.
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Yes it is true. Did you read the question?
On day one, one person is infected. On day two that one person infects two more people.
Now we have three people infected. On day three those three people infect six more people.
Now we have nine, [imath]3^{3-1}[/imath], people infected.
 
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