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

    Now just differentiate both sides again wrt x and multiply both sides by y.
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    (Original post by RDKGames)
    Have you found \frac{dy}{dx} 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?
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    (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 \frac{dy}{dx}=0 for an expression.

    You have y=x^{-x}, \forall x \in \{ \mathbb{R} : x>0 \} and you need to differentiate this. Begin by rewriting this as \ln(y)=\ln(x^{-x})=-x\ln(x) and proceed with implicit differentiation on the LHS for \frac{dy}{dx}
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    (Original post by RDKGames)
    Well the first part is asking for a stationary point. Stationary points are found when you set \frac{dy}{dx}=0 for an expression.

    You have y=x^{-x}, \forall x \in \{ \mathbb{R} : x>0 \} and you need to differentiate this. Begin by rewriting this as \ln(y)=\ln(x^{-x})=-x\ln(x) and proceed with implicit differentiation on the LHS for \frac{dy}{dx}
    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
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    (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 \frac{dy}{dx}=-y(1+\ln(x)) and differentiate it again to find \frac{d^2 x}{dy^2}. Then multiply both sides by y of the expression and tidy it up.

    For Q5 if you start your inductive step with  \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac  {1}{2})^{k+1} and up with 2-(k+3)(\frac{1}{2})^{k+1} then you've basically proved it. You can turn it into 2-[(k+1)+2](\frac{1}{2})^{k+1} for clarity which shows that if the statement is true for n=k then it is true for n=k+1 and you just need to write an ending statement to that.
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    (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 \frac{dy}{dx}=-y(1+\ln(x)) and differentiate it again to find \frac{d^2 x}{dy^2}. Then multiply both sides by y of the expression and tidy it up.

    For Q5 if you start your inductive step with  \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac  {1}{2})^{k+1} and up with 2-(k+3)(\frac{1}{2})^{k+1} then you've basically proved it. You can turn it into 2-[(k+1)+2](\frac{1}{2})^{k+1} for clarity which shows that if the statement is true for n=k then it is true for n=k+1 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 :/
     \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac  {1}{2})^{k+1}
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    (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 :/
     \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k+(k+1)(\frac  {1}{2})^{k+1}
    Right so when you do it for n=1 it works out.

    Now assume it works for n=k therefore out inductive assumption is that \displaystyle \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k

    Next we must test it for n=k+1

    So we have \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] = \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] + \underbrace{(k+1)(\frac{1}{2})^{  k+1}}_{(k+1)^{th} \text{term}} = \underbrace{2-(k+2)(\frac{1}{2})^k}_{\text{ inductive assumption}} + (k+1)(\frac{1}{2})^{k+1}
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    (Original post by RDKGames)
    Right so when you do it for n=1 it works out.

    Now assume it works for n=k therefore out inductive assumption is that \displaystyle \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] =2-(k+2)(\frac{1}{2})^k

    Next we must test it for n=k+1

    So we have \displaystyle \sum_{r=1}^{k+1} [r\cdot (\frac{1}{2})^r] = \sum_{r=1}^{k} [r\cdot (\frac{1}{2})^r] + \underbrace{(k+1)(\frac{1}{2})^{  k+1}}_{(k+1)^{th} \text{term}} = \underbrace{2-(k+2)(\frac{1}{2})^k}_{\text{ inductive assumption}} + (k+1)(\frac{1}{2})^{k+1}
    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?
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    (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 n=k+1 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 r=1 to r=k+1 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 n=k+1
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    (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  \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) = 2 - \displaystyle \left(k+2\right)\left(\frac{1}{2  } \right)^{k} .

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

     \displaystyle\sum_{r=1}^{k+1} \left( \left(\frac{1}{2}\right)^r r\right) =  \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) + \text{the }(k+1)^{\text{th}} \text{ term} ,

    which allows us to use the assumption by just substitution.
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    (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  \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) = 2 - \displaystyle \left(k+2\right)\left(\frac{1}{2  } \right)^{k} .

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

     \displaystyle\sum_{r=1}^{k+1} \left( \left(\frac{1}{2}\right)^r r\right) =  \displaystyle\sum_{r=1}^k \left( \left(\frac{1}{2}\right)^r r\right) + \text{the }(k+1)^{\text{th}} \text{ term} ,

    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?
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    (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
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    (Original post by crashMATHS)
    Yes, if I understand what you're saying correctly
    Thank you

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