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haar wavelet notes 1

haar wavelet notes 2

Please explain what limits of integration they took for solving $p_i(x)$ and $q_i(x)$.

like for 1st step of $p_i(x)$:

$$\int_\alpha^x 1 \, \mathrm dx = x-\alpha$$

but for 2nd step of $p_i(x)$:

$$\int_\gamma^x -1 \,\mathrm dx = \gamma-x$$

Why did they take this limit? Why did they skipp beta? Please explain! and for $q_i(x)$ as well...

Milind Hegde
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Intellectual_
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1 Answers1

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We want to obtain an formula for $p_i$ knowing that $p_i'=h_i$. Of course we use $$p_i(x)=p_i(x_0)+\int_{x_0}^x h_i(t)\,dt \tag1$$ where $x_0$ can be chosen as we wish (as long as we can find $p_i(x_0)$ easily). And we wish to choose $x_0$ to simplify integration: it's better not to have the transition point $\beta$ inside of the interval of integration.

So, when $x\in [\alpha,\beta)$, we pick $x_0=\alpha$ where $p_i(\alpha)=0$. But when $x\in [\beta,\gamma)$, we pick $x_0=\gamma$ where $p_i(\gamma)=\int_{\alpha}^\gamma h_i=0$. This simplifies integration: $p_i(x)=\int_{\gamma}^x -1\,dt = \gamma-x$.

When it comes to $q_i$, we have $$q_i(x)=q_i(x_0)+\int_{x_0}^x p_i(t)\,dt \tag2$$ When $x\in [\alpha,\beta)$, we pick $x_0=\alpha$ where $q_i(\alpha)=0$. When $x\in [\beta,\gamma)$, we pick $x_0=\gamma$ where $q_i(\gamma)=\int_{\alpha}^\gamma p_i=\frac12((\alpha-\beta)^2+(\beta-\gamma)^2)$.

ˈjuː.zɚ79365
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  • since x∈[β,γ), why did not we take ∫−1dt lower limit x and upper limit γ ?? – Intellectual_ Jun 26 '13 at 16:14
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    @Intellectual_ In formula (1), $x$ is the upper limit. If you want to have it as a lower limit, you should put $-$ sign in front of integral. But why do that? $\int_a^b$ makes perfect sense when $a>b$. – ˈjuː.zɚ79365 Jun 26 '13 at 23:30
  • thank you! i did understand pi(x) but still have queries about qi(x). i dont know how to write equations here so i made the following page to get help from you. Please answer my remaining queries too... https://docs.google.com/drawings/d/1AZLNP2zz1__9rK7npFmXU0W5O5sh1Ra3dJ2-hBhRBgs/edit?usp=sharing – Intellectual_ Jun 27 '13 at 07:56
  • $p_i (x)$ = $\int_\alpha^x 1 \ dx =x-\alpha, for [\alpha,\beta) $
    $p_i (x)$ = $\int_\gamma^x -1 \ dx =\gamma-x, for [\beta,\gamma)$,
    $p_i (x)$=0 elsewhere
    I understood these steps.
    Now for $q_i (x)$
    $q_i (x)$ = $\int_\alpha^x (x-\alpha) \ dx =(x-\alpha)^2 /2, for [\alpha,\beta)$,
    $q_i (x)$ = $\int_\alpha^\beta (x-\alpha)+ \int_\beta^x (\gamma-x)\ dx =[(x-\alpha)^2 +(\gamma-x)^2-(\beta-\gamma)^2]/2, for [\beta,\gamma)$,
    which is not same as calculated by the auther, means I am doing some mistake again in choosing the limits.     – Intellectual_ Jun 27 '13 at 10:04
  • and again
    $q_i (x)$ = $\int_\alpha^\beta (x-\alpha)+ \int_\beta^\gamma (\gamma-x)\ dx + \int_\gamma^x (\gamma-x)\ dx=[(x-\alpha)^2 +(\gamma-x)^2+(\beta-\gamma)^2]/2, for [\gamma,1)$,
    which is not same as required. please let me know where am I going wrong?     – Intellectual_ Jun 27 '13 at 10:05
  • ok, I found my mistakes, was not taking '-' out while integrating $(\gamma-x)$ – Intellectual_ Jun 29 '13 at 15:27