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# further pure 1 watch

1. http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
Can someone help me with 9b?
2. (Original post by English-help)
http://mathsathawthorn.pbworks.com/f/FP1Jun06.pdf
Can someone help me with 9b?
Have you found in terms of y and x??

Now just differentiate both sides again wrt x and multiply both sides by y.
3. (Original post by RDKGames)
Have you found in terms of y and x??

Now just differentiate both sides again wrt x and multiply both sides by y.
Tbh i dont even understand the question , what is it trying to tell me to find out and how should i start?
4. (Original post by English-help)
Tbh i dont even understand the question , what is it trying to tell me to find out and how should i start?
Well the first part is asking for a stationary point. Stationary points are found when you set for an expression.

You have and you need to differentiate this. Begin by rewriting this as and proceed with implicit differentiation on the LHS for
5. (Original post by RDKGames)
Well the first part is asking for a stationary point. Stationary points are found when you set for an expression.

You have and you need to differentiate this. Begin by rewriting this as and proceed with implicit differentiation on the LHS for
Ive done 9a! :/ I was talking about 9b?
Sorry , Well i did say 9b in the OP
http://mathsathawthorn.pbworks.com/f/FP1Jun07.pdf
I need help on q5^ Basically I have done n=1 and n=k , when i put n=k+1 inside the function , it gives me 2-(k+3)(1/2)^k+1
Idk what to do next though
6. (Original post by English-help)
Ive done 9a! :/ I was talking about 9b?
Sorry , Well i did say 9b in the OP
http://mathsathawthorn.pbworks.com/f/FP1Jun07.pdf
I need help on q5^ Basically I have done n=1 and n=k , when i put n=k+1 inside the function , it gives me 2-(k+3)(1/2)^k+1
Idk what to do next though
When you said "Tbh i dont even understand the question" I assumed you simply didn't know where to start on the question altogether. For 9b, as I said, just take your and differentiate it again to find . Then multiply both sides by of the expression and tidy it up.

For Q5 if you start your inductive step with and up with then you've basically proved it. You can turn it into for clarity which shows that if the statement is true for then it is true for and you just need to write an ending statement to that.
7. (Original post by RDKGames)
When you said "Tbh i dont even understand the question" I assumed you simply didn't know where to start on the question altogether. For 9b, as I said, just take your and differentiate it again to find . Then multiply both sides by of the expression and tidy it up.

For Q5 if you start your inductive step with and up with then you've basically proved it. You can turn it into for clarity which shows that if the statement is true for then it is true for and you just need to write an ending statement to that.
But do you know , how did you get this step when doing n=k+1 , thats what i dont understand :/
8. (Original post by English-help)
But do you know , how did you get this step when doing n=k+1 , thats what i dont understand :/
Right so when you do it for it works out.

Now assume it works for therefore out inductive assumption is that

Next we must test it for

So we have
9. (Original post by RDKGames)
Right so when you do it for it works out.

Now assume it works for therefore out inductive assumption is that

Next we must test it for

So we have
So you get the get the Inductive assumption from n=k and then do n=k=1 which comes from the LHS? + then add both of them?
10. (Original post by English-help)
So you get the get the Inductive assumption from n=k
Yes we need to establish an assumption before we can proceed with the proof.

and then do n=k+1 which comes from the LHS? + then add both of them?
You test for and to do so you need to make use of the assumption you made earlier. The only way to do this is to split the sum from to into 2 different terms where one can be replaced by the assumption (the one which goes from r=1 to r=k) and the other is whatever is left over to balance the sum (the (k+1)th term). So then you just add the two things and should end up with an expression for
11. (Original post by English-help)
So you get the get the Inductive assumption from n=k and then do n=k=1 which comes from the LHS? + then add both of them?
To add to what RDK games has said, the whole structure of a proof by induction is to make some use of your assumption in the inductive step - that's where the real power of this type of proof comes from.

In this case, you have assumed that .

So a nice way in which we can use our assumption here is if we partition the sum in our inductive step:

,

which allows us to use the assumption by just substitution.
12. (Original post by crashMATHS)
To add to what RDK games has said, the whole structure of a proof by induction is to make some use of your assumption in the inductive step - that's where the real power of this type of proof comes from.

In this case, you have assumed that .

So a nice way in which we can use our assumption here is if we partition the sum in our inductive step:

,

which allows us to use the assumption by just substitution.
So you get the K+1 term when you put k+1 into the left hand side basically and the K term from when you do n=k? right?
13. (Original post by English-help)
So you get the K+1 term when you put k+1 into the left hand side basically and the K term from when you do n=k? right?
Yes, if I understand what you're saying correctly
14. (Original post by crashMATHS)
Yes, if I understand what you're saying correctly
Thank you

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