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Math Help - a few random questions.

  1. #1
    Junior Member pearlyc's Avatar
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    a few random questions.

    a few random questions :

    1. Given :
    1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}

    (a) Prove this results from first principles.
    2. Show from first principles that,
    \sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}


    Hence give an expression of \sum^{n-1}_{i=0} x^i
    3. Let
    g(t) = exp \{ \alpha + \beta t \}
    Given that g(10) = 8.1882 and g(20) = 60.3403, calculate g(15).

    4. Consider a function f such that f(11) = 1.234 and f(12)=2.345. Assuming that the function is linear over the interval from 11 to 12, calculate f(11.36).

    Someone would please kindly guide me on how to do the questions! Thanks in advance!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by pearlyc View Post
    a few random questions :

    1. Given :
    1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}

    (a) Prove this results from first principles
    This depends on what you mean by first principles. Does induction count?

    What about this:

    Let:

    S(n)=1 + 2 + 3 + ... + n

    Then:

    (\Delta S)(n)=S(n)-S(n-1)=n,\ \ \ n\ge 2

     <br />
(\Delta^2 S)(n)=(\Delta S)(n)-(\Delta S)(n-1)=1, \ \ \ n\ge 3<br />

    That is the second differences of S(n) are constant, which means that S(n) is a quadratic in n.

    So let:

     <br />
S(n)=a n^2 + b n+c<br />

    Now:

    S(1)=a+b+c=1
    S(2)=4a+2b+c=3
    S(3)=9a+3b+c=6

    Thus we have a set of three simultaneous equations in a, b and c which we solve to find the quadratic for S(n).

    RonL
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  3. #3
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    Hello, pearlyc!

    3. Let g(t) \:= \:e^{\alpha +\beta\,t}

    Given that: . g(10) = 8.1882\,\text{ and }\,g(20) = 60.3403

    . . calculate g(15).

    We have: . \begin{array}{cccccc}g(10)\:=\:8.1882 & \Rightarrow & e^{(\alpha+10\beta)} &=& 8.1882 & {\color{blue}[1]} \\<br />
g(20)\:=\:60.3402 & \Rightarrow & e^{(\alpha+20\beta)} &=& 60.3402 & {\color{blue}[2]}\end{array}

    Divide [2] by [1]: . \frac{e^{(\alpha + 20\beta)}}{e^{(\alpha+10\beta)}} \;=\;\frac{60.3402}{8.1882} \quad\Rightarrow\quad e^{10\beta} \:=\:7.369165384

    Then: . 10\beta \:=\:\ln(7.369165384)

    . . Hence: . \beta \:=\:0.199730445 \quad\Rightarrow\quad\boxed{\beta \:\approx\:0.2}

    So far: . g(t) \;=\;e^{(\alpha+0.2t)}


    Then [1] becomes: . g(10) \;=\;e^{\alpha +2} \;=\;8.1882 \quad\Rightarrow\quad \alpha + 2 \:=\:\ln(8.1882)

    . . Hence: . \alpha \:=\:0.102694093\quad\Rightarrow\quad \boxed{\alpha \:\approx\:0.1}


    And we have: . g(t) \;=\;e^{0.1 +0.2t}

    Therefore: . g(15)\;=\;e^{0.1 + 3} \:=\:e^{3.1} \:=\:22.19795128 \:\approx\:22.2

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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by pearlyc View Post
    2. Show from first principles that,
    \sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}


    Hence give an expression of \sum^{n-1}_{i=0} x^i
    (1-x)\sum^n_{i=1} x^i = \sum_{i=1}^n (x^i-x^{i+1})=(x^1-x^2) +(x^2-x^3) + ... +(x^{n-1}-x^n)+(x^n-x^{n+1})

    Collecting like powers:

    (1-x)\sum^n_{i=1} x^i = x^1-x^{n+1}=x(1-x^n)

    RonL
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