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| visits | member for | 2 years, 6 months |
| seen | Mar 10 at 19:08 | |
| stats | profile views | 133 |
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Nov 23 |
comment |
Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ ok so what i compute is: $$M_{n+m} = M_n*M_m = \left(\begin{array}{ccc} F_{n+1} & F_n \\\ F_n & F_{n-1} \end{array}\right)*\left(\begin{array}{ccc} F_{m+1} & F_m \\\ F_m & F_{m-1} \end{array}\right) = \left(\begin{array}{ccc} F_{n+1}F_{m+1} + F_nF_m & F_{n+1}F_m + F_nF_{m-1} \\\ F_nF_{m+1} + F_{n-1}F_m & F_{n}F_{m} + F_{n-1}F_{m-1} \end{array}\right)$$ which as you said shows the desired expression in the bottom left(and i assume equivalent) in the top right. is this method something one can usually use with sequences? i guess i just don't know how it works in general... |
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Nov 23 |
comment |
Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ i'm sorry, but i'm having trouble figuring out how to use that to simplify.... |
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Nov 23 |
awarded | Commentator |
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Nov 23 |
comment |
Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ am i interpreting this properly if i write: $\frac{\Phi^{n+m} - (1-\Phi)^{n+m}}{\sqrt{5}} = \frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}} * \frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}} + \frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}} * \frac{\Phi^{1+m} - (1-\Phi)^{1+m}}{\sqrt{5}}$ $\frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}}(\frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}} + \frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}} * F_1)$ $\frac{\Phi^{m} - (1-\Phi)^{m}}{\sqrt{5}}(\frac{\Phi^{n-1} - (1-\Phi)^{n-1}}{\sqrt{5}}+\frac{\Phi^{n} - (1-\Phi)^{n}}{\sqrt{5}})$ ? what can i do at the end? |
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Nov 23 |
asked | Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ |
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Nov 21 |
comment |
Are there linear mappings for the following vectors, if so, what are they? @Arturo thank you so much for your help!!(same goes for everyone else who contributed of course..). Just one minor thing: i believe that where you wrote the equation for $f(0,1,0)$ the vector $\frac{1}{4}(1,0,2)$ should have been $(0,-2,1)$, correct? Also i don't totally 'get' why we're allowed to map $(1,0,0)$ to 'anything', but that is most likely beyond the scope of this and will hopefully become clear to me soon anyway. other than that cannot emphasize enough how much i'm learning and how helpful this has been! |
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Nov 20 |
awarded | Scholar |
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Nov 20 |
accepted | Are there linear mappings for the following vectors, if so, what are they? |
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Nov 20 |
comment |
Are there linear mappings for the following vectors, if so, what are they? $\begin{array}{rrrr} x & +0y& + 2z &=a \\ x& +2y &+0z &=b \\x& + y& +z&=c \end{array}$ i then use the following equalities and substitutions: $x=-2z+a$ $y=\frac{1}{2} (b-x)$ $z=-(-2z+a)-[\frac{1}{2}(b-x)]+c$ $z=2z-a-\frac{1}{2}b+\frac{1}{2}x+c$ $z=2z-a-\frac{1}{2}b+\frac{1}{2}(-2z+a)+c$ $z=2z-z-\frac{1}{2}a-\frac{1}{2}b+c$ $z=z-\frac{1}{2}a-\frac{1}{2}b+c$ $0=-\frac{1}{2}a-\frac{1}{2}b+c$ but would this have been valid and sufficient? (i don't know if this actually shows that the system is inconsistent.....) |
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Nov 20 |
comment |
Are there linear mappings for the following vectors, if so, what are they? @Arturo i think i'm understanding what you're saying. definitely the part about concluding things incorrectly. my only problem is that i am having trouble understanding the general approach to such a problem... before your response i was about to post: linear combination: $(a,b,c) = x \left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right) + y \left (\begin{array}{c} 0\\2\\1 \end{array} \right) + z \left ( \begin{array}{c} 2\\0\\1 \end{array} \right) = \left ( \begin{array}{c} x + 2z \\ x + 2y \\ x + y + z \end{array} \right)$ |
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Nov 20 |
comment |
Are there linear mappings for the following vectors, if so, what are they? ok i'm sorry for being so clueless, but is this in the right direction? as a linear combination i have: $(a,b,c) = x \left( \begin{array}{c} 1\\ 1\\ 1 \end{array} \right) + y \left (\begin{array}{c} 0\\2\\1 \end{array} \right) + z \left ( \begin{array}{c} 2\\0\\1 \end{array} \right) = \left ( \begin{array}{c} x + 2z \\ x + 2y \\ x + y + z \end{array} \right)$ |
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Nov 20 |
awarded | Student |
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Nov 20 |
asked | Are there linear mappings for the following vectors, if so, what are they? |
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Nov 20 |
comment |
which of the following are linear mappings? or for (a) would it simply be enough to say that since $\varphi(0) \neq 0$ it does not satisfy the condition of scalar multiplication and is therefore not a linear map? |
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Nov 20 |
comment |
which of the following are linear mappings? ok cool so (b) is taken care of.. does that mean i got (a) wrong? any other clues for (c) and (d)? i am still having difficulty understanding them |
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Nov 20 |
comment |
which of the following are linear mappings? p.s i really appreciate that you responded so thoughfully and quickly :) |
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Nov 20 |
comment |
which of the following are linear mappings? with (b) i'm not sure if i totally understand the concept, is it that the function of a function is equalto the function's value at one? in this case i would like to simply write $\varphi(f+g)=(f+g)(1)=f(1)+g(1)=\varphi(f)+\varphi(g) and \lambda\varphi(f) = \lambda f(1) = \varphi(\lambda f) \Rightarrow$ (b) is a linear mapping?? |
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Nov 20 |
comment |
which of the following are linear mappings? Hmm so if I understand that correctly (a) is clearly not a linear mapping since: $\varphi(x+y)=(1-x-y,(1-x-y)^{2})\neq(2-x-y,(1-x)^{2}+(1-y)^{2})=\varphi(x)+\varphi(y)$ is that correct? |
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Nov 19 |
asked | which of the following are linear mappings? |
