Here is the solution with the question. https://www.quora.com/profile/Bravewarrior/Need-help-with-part-c

How would I get to 0.01 being the value to substitute into the expansion? Thanks in adavnce!

How would I get to 0.01 being the value to substitute into the expansion? Thanks in adavnce!

Original post by pigeonwarrior

Here is the solution with the question. https://www.quora.com/profile/Bravewarrior/Need-help-with-part-c

How would I get to 0.01 being the value to substitute into the expansion? Thanks in adavnce!

How would I get to 0.01 being the value to substitute into the expansion? Thanks in adavnce!

You want to find 101/103 using (1+x)/(1+3x).

Can you not just sub x=0.01 in and see if / why it works?

Original post by mqb2766

You want to find 101/103 using (1+x)/(1+3x).

Can you not just sub x=0.01 in and see if / why it works?

Can you not just sub x=0.01 in and see if / why it works?

Helloo I did sub in 0.01 but my question is how am I supposed to get to 0.01 in the first place?

Original post by pigeonwarrior

Helloo I did sub in 0.01 but my question is how am I supposed to get to 0.01 in the first place?

For other problems like this, how do they work out the x to use? What properties should it have and how do they apply here?

Original post by mqb2766

For other problems like this, how do they work out the x to use? What properties should it have and how do they apply here?

icl i never learnt why they sub in 'xyz' values into x but from the questions ive done its usually always 0.01 or 0.1

Original post by mqb2766

For other problems like this, how do they work out the x to use? What properties should it have and how do they apply here?

To be completely honest this is the first question in this chapter where they ask me to choose a suitable value otherwise the question normally gives the value to sub in from what I've seen so far 😭😅

Original post by anonebony

icl i never learnt why they sub in 'xyz' values into x but from the questions ive done its usually always 0.01 or 0.1

Thankyouu I will probably just memorise this and hope for the best tbh...😭

Original post by pigeonwarrior

Helloo I did sub in 0.01 but my question is how am I supposed to get to 0.01 in the first place?

Can you not see a link between 0.01 and 101?

Original post by pigeonwarrior

To be completely honest this is the first question in this chapter where they ask me to choose a suitable value otherwise the question normally gives the value to sub in from what I've seen so far 😭😅

Wbf, if youre just subbing numbers into formulae without understanding why youll come unstuck in exam questions.

For this one, theres only one possibility as the numerator and denominator can be multipled (divided) by an arbitrary constant "k" and the fraction remains unchanged. So here

101 = k(1+x)

103 = k(1+3x)

Solving gives x=0.01 and k=100.

This is a bit unusual and normally the choice of x isnt unique, but there should be an "obvious" one. Obvious would mean something like

•

|x| << 1 as the smaller it is, the less terms you need to compute to get a given accuracy and you need |x|<1/3 here to guarantee convergence anyways.

•

x ~ 10^(-n), as this both makes the computation easy to evaluate and easy to reason about the number of decimal places it will be accurate to.

Original post by pigeonwarrior

Thankyouu I will probably just memorise this and hope for the best tbh...😭

oh i just clocked how to do it lol, hopefully its not incorrect

take the initial equation and equate it to the fraction/value they provide you with:

(1-x)/(1-3x)=101/103

solve to find x

103(1-x)=103(1-3x)

blah blah and then u get 0.01

Original post by Muttley79

Can you not see a link between 0.01 and 101?

I see it now!

Original post by mqb2766

Wbf, if youre just subbing numbers into formulae without understanding why youll come unstuck in exam questions.

For this one, theres only one possibility as the numerator and denominator can be multipled (divided) by an arbitrary constant "k" and the fraction remains unchanged. So here

101 = k(1+x)

103 = k(1+3x)

Solving gives x=0.01 and k=100.

This is a bit unusual and normally the choice of x isnt unique, but there should be an "obvious" one. Obvious would mean something like

In practice it could be related to square .... so its not always 0.1 or 0.01, but thinking like that is a decent place to start.

For this one, theres only one possibility as the numerator and denominator can be multipled (divided) by an arbitrary constant "k" and the fraction remains unchanged. So here

101 = k(1+x)

103 = k(1+3x)

Solving gives x=0.01 and k=100.

This is a bit unusual and normally the choice of x isnt unique, but there should be an "obvious" one. Obvious would mean something like

•

|x| << 1 as the smaller it is, the less terms you need to compute to get a given accuracy and you need |x|<1/3 here to guarantee convergence anyways.

•

x ~ 10^(-n), as this both makes the computation easy to evaluate and easy to reason about the number of decimal places it will be accurate to.

Thank you so so much, this makes much more sense now! Thank you!!!

Original post by anonebony

oh i just clocked how to do it lol, hopefully its not incorrect

take the initial equation and equate it to the fraction/value they provide you with:

(1-x)/(1-3x)=101/103

solve to find x

103(1-x)=103(1-3x)

blah blah and then u get 0.01

take the initial equation and equate it to the fraction/value they provide you with:

(1-x)/(1-3x)=101/103

solve to find x

103(1-x)=103(1-3x)

blah blah and then u get 0.01

It looks extremely right to me, thank youu for this!!!

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